$limlimits_{xto 0} frac{tan x - sin x}{x^3}$?












8














$$lim_{xto 0} frac{tan x - sin x}{x^3}$$



Solution



begin{align}lim_{xto 0} frac{tan x - sin x}{x^3}&=\&=lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}\
&= lim_{xto 0}frac{tan x}{x}lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{sin x}{x}lim_{xto 0} frac{1}{x^2}\&= lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{1}{x^2}\
&= lim_{xto 0} frac{1}{x^2} -frac{1}{x^2}\&=0 end{align}



But the answer is $dfrac{1}{2}$ by L'Hopital's Rule.










share|cite|improve this question




















  • 8




    $infty -infty = ?$
    – Math Lover
    Nov 21 at 17:33










  • edited it, $lim_{xto 0} space 0 = 0$
    – Raghav
    Nov 21 at 17:37












  • The same mistake applies regardless. You can’t break down a limit like that if the any of the individual limits are undefined/don’t exist.
    – KM101
    Nov 21 at 17:38












  • But if we have $ lim_{xto 1} frac{x}{x-1} - frac{1}{x-1}$, It will be indeterminate form if we put X=1 in it but we get 1 by adding them.
    – Raghav
    Nov 21 at 17:44






  • 4




    Please re-write your title to be informative. Someone else with the exact same problem would never find this solution based on its title.
    – David G. Stork
    Nov 21 at 19:36
















8














$$lim_{xto 0} frac{tan x - sin x}{x^3}$$



Solution



begin{align}lim_{xto 0} frac{tan x - sin x}{x^3}&=\&=lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}\
&= lim_{xto 0}frac{tan x}{x}lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{sin x}{x}lim_{xto 0} frac{1}{x^2}\&= lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{1}{x^2}\
&= lim_{xto 0} frac{1}{x^2} -frac{1}{x^2}\&=0 end{align}



But the answer is $dfrac{1}{2}$ by L'Hopital's Rule.










share|cite|improve this question




















  • 8




    $infty -infty = ?$
    – Math Lover
    Nov 21 at 17:33










  • edited it, $lim_{xto 0} space 0 = 0$
    – Raghav
    Nov 21 at 17:37












  • The same mistake applies regardless. You can’t break down a limit like that if the any of the individual limits are undefined/don’t exist.
    – KM101
    Nov 21 at 17:38












  • But if we have $ lim_{xto 1} frac{x}{x-1} - frac{1}{x-1}$, It will be indeterminate form if we put X=1 in it but we get 1 by adding them.
    – Raghav
    Nov 21 at 17:44






  • 4




    Please re-write your title to be informative. Someone else with the exact same problem would never find this solution based on its title.
    – David G. Stork
    Nov 21 at 19:36














8












8








8


2





$$lim_{xto 0} frac{tan x - sin x}{x^3}$$



Solution



begin{align}lim_{xto 0} frac{tan x - sin x}{x^3}&=\&=lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}\
&= lim_{xto 0}frac{tan x}{x}lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{sin x}{x}lim_{xto 0} frac{1}{x^2}\&= lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{1}{x^2}\
&= lim_{xto 0} frac{1}{x^2} -frac{1}{x^2}\&=0 end{align}



But the answer is $dfrac{1}{2}$ by L'Hopital's Rule.










share|cite|improve this question















$$lim_{xto 0} frac{tan x - sin x}{x^3}$$



Solution



begin{align}lim_{xto 0} frac{tan x - sin x}{x^3}&=\&=lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}\
&= lim_{xto 0}frac{tan x}{x}lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{sin x}{x}lim_{xto 0} frac{1}{x^2}\&= lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{1}{x^2}\
&= lim_{xto 0} frac{1}{x^2} -frac{1}{x^2}\&=0 end{align}



But the answer is $dfrac{1}{2}$ by L'Hopital's Rule.







calculus limits proof-verification limits-without-lhopital






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 11:41









amWhy

191k28224439




191k28224439










asked Nov 21 at 17:31









Raghav

557




557








  • 8




    $infty -infty = ?$
    – Math Lover
    Nov 21 at 17:33










  • edited it, $lim_{xto 0} space 0 = 0$
    – Raghav
    Nov 21 at 17:37












  • The same mistake applies regardless. You can’t break down a limit like that if the any of the individual limits are undefined/don’t exist.
    – KM101
    Nov 21 at 17:38












  • But if we have $ lim_{xto 1} frac{x}{x-1} - frac{1}{x-1}$, It will be indeterminate form if we put X=1 in it but we get 1 by adding them.
    – Raghav
    Nov 21 at 17:44






  • 4




    Please re-write your title to be informative. Someone else with the exact same problem would never find this solution based on its title.
    – David G. Stork
    Nov 21 at 19:36














  • 8




    $infty -infty = ?$
    – Math Lover
    Nov 21 at 17:33










  • edited it, $lim_{xto 0} space 0 = 0$
    – Raghav
    Nov 21 at 17:37












  • The same mistake applies regardless. You can’t break down a limit like that if the any of the individual limits are undefined/don’t exist.
    – KM101
    Nov 21 at 17:38












  • But if we have $ lim_{xto 1} frac{x}{x-1} - frac{1}{x-1}$, It will be indeterminate form if we put X=1 in it but we get 1 by adding them.
    – Raghav
    Nov 21 at 17:44






  • 4




    Please re-write your title to be informative. Someone else with the exact same problem would never find this solution based on its title.
    – David G. Stork
    Nov 21 at 19:36








8




8




$infty -infty = ?$
– Math Lover
Nov 21 at 17:33




$infty -infty = ?$
– Math Lover
Nov 21 at 17:33












edited it, $lim_{xto 0} space 0 = 0$
– Raghav
Nov 21 at 17:37






edited it, $lim_{xto 0} space 0 = 0$
– Raghav
Nov 21 at 17:37














The same mistake applies regardless. You can’t break down a limit like that if the any of the individual limits are undefined/don’t exist.
– KM101
Nov 21 at 17:38






The same mistake applies regardless. You can’t break down a limit like that if the any of the individual limits are undefined/don’t exist.
– KM101
Nov 21 at 17:38














But if we have $ lim_{xto 1} frac{x}{x-1} - frac{1}{x-1}$, It will be indeterminate form if we put X=1 in it but we get 1 by adding them.
– Raghav
Nov 21 at 17:44




But if we have $ lim_{xto 1} frac{x}{x-1} - frac{1}{x-1}$, It will be indeterminate form if we put X=1 in it but we get 1 by adding them.
– Raghav
Nov 21 at 17:44




4




4




Please re-write your title to be informative. Someone else with the exact same problem would never find this solution based on its title.
– David G. Stork
Nov 21 at 19:36




Please re-write your title to be informative. Someone else with the exact same problem would never find this solution based on its title.
– David G. Stork
Nov 21 at 19:36










5 Answers
5






active

oldest

votes


















12














Your problem arises from the fact that you used $color{red}{lim_limits{x to 0} frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $color{red}{infty-infty}$...



Only split an initial limit into a product if the individual limits are defined.






share|cite|improve this answer





















  • Can't I obtain 0 by subtracting $frac{1}{x^2}$ by $frac{1}{x^2}$?
    – Raghav
    Nov 21 at 17:39






  • 1




    You can’t break down a limit into a product if the individual limits aren’t defined. That’s where your error arose.
    – KM101
    Nov 21 at 17:40












  • Ohh, ok. So the the functions need to exist at the limit to be able to break down. Thank you very much man!
    – Raghav
    Nov 21 at 17:46










  • Exactly! (No problem.)
    – KM101
    Nov 21 at 17:47










  • @Raghav - The function does not need to exist, but the limit does. If two of the three limits $lim f, lim g, lim (f + g)$ are known to exist (as a finite number), then it is guaranteed that the third also exists and the relationshil $$lim f + lim g = lim (f+g)$$ holds. But when only one of the limits is known to exist, there is no reason that the formula has to be true.
    – Paul Sinclair
    Nov 22 at 2:58



















28














This is just another way of saying what the others told you.



$$lim_{xto 0} frac{tan x - sin x}{x^3}
ne lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}$$



The theorem is
IF $displaystyle lim_{xto 0}f(x) = L$
and $displaystyle lim_{xto 0}g(x)=M$, where $M, N in mathbb R$,
THEN $displaystyle lim_{xto 0}(f(x)-g(x))=L-M$



But, since $displaystyle lim_{xto 0} frac{tan x}{x^3} = lim_{xto 0} frac{sin x}{x^3} = infty$, then the theorem does not apply.



This limit can be evaluated without resorting to L'Hospital.



begin{align}
frac{tan x - sin x}{x^3}
&= frac{frac{sin x}{cos x} - sin x}{x^3} \
&= frac{sin x - sin x cos x}{x^3 cos x} \
&= frac{1}{cos x} cdotfrac{sin x}{x} cdot frac{1 - cos x}{x^2} \
&= frac{1}{cos x} cdotfrac{sin x}{x}
cdot frac{2sin^2(frac 12x)}{x^2} \
&= frac{1}{cos x} cdotfrac{sin x}{x}
cdot frac 12 cdot left(frac{sin frac x2}{frac x2}right)^2 \
end{align}



which approaches $dfrac 12$ as $x$ approaches $0$.






share|cite|improve this answer



















  • 1




    @Steven where did I do $infty - infty$
    – Akash Roy
    Nov 22 at 13:05






  • 1




    Your way is correct.
    – KM101
    Nov 22 at 13:51






  • 1




    @AkashRoy Yes what you did is correct. I was looking at someone elses answer by mistake.
    – steven gregory
    Nov 22 at 14:14








  • 1




    I didn't know we needed to go to Le Hospital to solve limits :)
    – Abraham Zhang
    Nov 22 at 22:36






  • 1




    @stevengregory Check your spelling of "L'Hôpital" and then re-read my comment.
    – Abraham Zhang
    Nov 23 at 1:07



















13














I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ limlimits_{x to 0}big( f(x) - g(x)big)$ is not always equal to $ limlimits_{x to 0} f(x) - limlimits_{x to 0} g(x)$.






share|cite|improve this answer



















  • 3




    This is the most concise correct answer. The obvious example is to pick any function $f:mathbb Rtomathbb R$ such that $limlimits_{xto 0}$ does not exist. Then $limlimits_{xto 0}big( f(x)-f(x)big)$ exists and equals $0$, but it is not equal to $limlimits_{xto 0}f(x) + limlimits_{xto 0}-f(x)$ since neither one of the latter two limits exists.
    – MPW
    Nov 21 at 18:30










  • There has to be further problems with the work, as splitting works when there's a finite limit. If the OP had worked through the problem correctly, they should have found that the separate limits were indeterminate. The issue is not just splitting, but then recombining them later to cancel out the indeterminate.
    – Acccumulation
    Nov 21 at 19:57










  • Is my answer okay?
    – Akash Roy
    Nov 22 at 8:00



















3














Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.



We have



$$lim_{x to 0} tan x= frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . .$$



$$lim_{x to 0} sin x= frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .$$



Therefore expression turns to,



$$lim_{x to 0} frac{frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . . - (frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .)}{x^{3}}$$



Cancel the $x$ and then enforce the limit after dividing the numerator by $x^{3}$ . The expression simplifies to the calculation of sum of $frac{1}{3}$ and $frac{1}{6}$ which is $color{red} {frac{1}{2}}$






share|cite|improve this answer























  • Cancel the x/1 term first and then divide the expression by $x^{3}$ then enforce the limit which is x tends to 0 .
    – Akash Roy
    Nov 22 at 5:33












  • Good solution, however I would use truncation (with big-$O$) rather than those dots. And then you don't even need the $x^5$ terms, so you can truncate at $O(x^5)$.
    – Jean-Claude Arbaut
    Nov 22 at 12:51










  • Ok sir Jean Claude Arbaut . Thanks for your words.
    – Akash Roy
    Nov 22 at 13:03










  • That's a nice solution.
    – Raghav
    Nov 22 at 16:24










  • Thanks @Raghav bro.
    – Akash Roy
    Nov 22 at 16:39



















2














Don't try to take the limit of each seperately & then take the difference - you just get the difference between two infinities! Express $sin$ & $tan$ as Taylor series - each has first term in $theta^1$ with coefficient 1, so in the difference it drops out. If you plot $sintheta-tantheta$ it looks like a cubic at the origin. Then if you divide that series by $theta^3$, & you get a series with an initial term in $theta^0$, ie a constant term. (This is shown in plots: if you plot that curve just described, ÷by $theta^3$, it begins somewhere along the y -axis instead of at the origin.) This is then all that is left as $thetarightarrow 0$. That's equivalent to tracing the plot I have just described in parenthesis to its point of intersection with the y -axis.



To actually get the answer immediately you just subtract the coefficient for $theta^3$ in the series for $sintheta$ from that in that for $tantheta$, & you get 1/3 - -1/6 = 1/2.






share|cite|improve this answer























    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008071%2flim-limits-x-to-0-frac-tan-x-sin-xx3%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    5 Answers
    5






    active

    oldest

    votes








    5 Answers
    5






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    12














    Your problem arises from the fact that you used $color{red}{lim_limits{x to 0} frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $color{red}{infty-infty}$...



    Only split an initial limit into a product if the individual limits are defined.






    share|cite|improve this answer





















    • Can't I obtain 0 by subtracting $frac{1}{x^2}$ by $frac{1}{x^2}$?
      – Raghav
      Nov 21 at 17:39






    • 1




      You can’t break down a limit into a product if the individual limits aren’t defined. That’s where your error arose.
      – KM101
      Nov 21 at 17:40












    • Ohh, ok. So the the functions need to exist at the limit to be able to break down. Thank you very much man!
      – Raghav
      Nov 21 at 17:46










    • Exactly! (No problem.)
      – KM101
      Nov 21 at 17:47










    • @Raghav - The function does not need to exist, but the limit does. If two of the three limits $lim f, lim g, lim (f + g)$ are known to exist (as a finite number), then it is guaranteed that the third also exists and the relationshil $$lim f + lim g = lim (f+g)$$ holds. But when only one of the limits is known to exist, there is no reason that the formula has to be true.
      – Paul Sinclair
      Nov 22 at 2:58
















    12














    Your problem arises from the fact that you used $color{red}{lim_limits{x to 0} frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $color{red}{infty-infty}$...



    Only split an initial limit into a product if the individual limits are defined.






    share|cite|improve this answer





















    • Can't I obtain 0 by subtracting $frac{1}{x^2}$ by $frac{1}{x^2}$?
      – Raghav
      Nov 21 at 17:39






    • 1




      You can’t break down a limit into a product if the individual limits aren’t defined. That’s where your error arose.
      – KM101
      Nov 21 at 17:40












    • Ohh, ok. So the the functions need to exist at the limit to be able to break down. Thank you very much man!
      – Raghav
      Nov 21 at 17:46










    • Exactly! (No problem.)
      – KM101
      Nov 21 at 17:47










    • @Raghav - The function does not need to exist, but the limit does. If two of the three limits $lim f, lim g, lim (f + g)$ are known to exist (as a finite number), then it is guaranteed that the third also exists and the relationshil $$lim f + lim g = lim (f+g)$$ holds. But when only one of the limits is known to exist, there is no reason that the formula has to be true.
      – Paul Sinclair
      Nov 22 at 2:58














    12












    12








    12






    Your problem arises from the fact that you used $color{red}{lim_limits{x to 0} frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $color{red}{infty-infty}$...



    Only split an initial limit into a product if the individual limits are defined.






    share|cite|improve this answer












    Your problem arises from the fact that you used $color{red}{lim_limits{x to 0} frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $color{red}{infty-infty}$...



    Only split an initial limit into a product if the individual limits are defined.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 21 at 17:37









    KM101

    3,970417




    3,970417












    • Can't I obtain 0 by subtracting $frac{1}{x^2}$ by $frac{1}{x^2}$?
      – Raghav
      Nov 21 at 17:39






    • 1




      You can’t break down a limit into a product if the individual limits aren’t defined. That’s where your error arose.
      – KM101
      Nov 21 at 17:40












    • Ohh, ok. So the the functions need to exist at the limit to be able to break down. Thank you very much man!
      – Raghav
      Nov 21 at 17:46










    • Exactly! (No problem.)
      – KM101
      Nov 21 at 17:47










    • @Raghav - The function does not need to exist, but the limit does. If two of the three limits $lim f, lim g, lim (f + g)$ are known to exist (as a finite number), then it is guaranteed that the third also exists and the relationshil $$lim f + lim g = lim (f+g)$$ holds. But when only one of the limits is known to exist, there is no reason that the formula has to be true.
      – Paul Sinclair
      Nov 22 at 2:58


















    • Can't I obtain 0 by subtracting $frac{1}{x^2}$ by $frac{1}{x^2}$?
      – Raghav
      Nov 21 at 17:39






    • 1




      You can’t break down a limit into a product if the individual limits aren’t defined. That’s where your error arose.
      – KM101
      Nov 21 at 17:40












    • Ohh, ok. So the the functions need to exist at the limit to be able to break down. Thank you very much man!
      – Raghav
      Nov 21 at 17:46










    • Exactly! (No problem.)
      – KM101
      Nov 21 at 17:47










    • @Raghav - The function does not need to exist, but the limit does. If two of the three limits $lim f, lim g, lim (f + g)$ are known to exist (as a finite number), then it is guaranteed that the third also exists and the relationshil $$lim f + lim g = lim (f+g)$$ holds. But when only one of the limits is known to exist, there is no reason that the formula has to be true.
      – Paul Sinclair
      Nov 22 at 2:58
















    Can't I obtain 0 by subtracting $frac{1}{x^2}$ by $frac{1}{x^2}$?
    – Raghav
    Nov 21 at 17:39




    Can't I obtain 0 by subtracting $frac{1}{x^2}$ by $frac{1}{x^2}$?
    – Raghav
    Nov 21 at 17:39




    1




    1




    You can’t break down a limit into a product if the individual limits aren’t defined. That’s where your error arose.
    – KM101
    Nov 21 at 17:40






    You can’t break down a limit into a product if the individual limits aren’t defined. That’s where your error arose.
    – KM101
    Nov 21 at 17:40














    Ohh, ok. So the the functions need to exist at the limit to be able to break down. Thank you very much man!
    – Raghav
    Nov 21 at 17:46




    Ohh, ok. So the the functions need to exist at the limit to be able to break down. Thank you very much man!
    – Raghav
    Nov 21 at 17:46












    Exactly! (No problem.)
    – KM101
    Nov 21 at 17:47




    Exactly! (No problem.)
    – KM101
    Nov 21 at 17:47












    @Raghav - The function does not need to exist, but the limit does. If two of the three limits $lim f, lim g, lim (f + g)$ are known to exist (as a finite number), then it is guaranteed that the third also exists and the relationshil $$lim f + lim g = lim (f+g)$$ holds. But when only one of the limits is known to exist, there is no reason that the formula has to be true.
    – Paul Sinclair
    Nov 22 at 2:58




    @Raghav - The function does not need to exist, but the limit does. If two of the three limits $lim f, lim g, lim (f + g)$ are known to exist (as a finite number), then it is guaranteed that the third also exists and the relationshil $$lim f + lim g = lim (f+g)$$ holds. But when only one of the limits is known to exist, there is no reason that the formula has to be true.
    – Paul Sinclair
    Nov 22 at 2:58











    28














    This is just another way of saying what the others told you.



    $$lim_{xto 0} frac{tan x - sin x}{x^3}
    ne lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}$$



    The theorem is
    IF $displaystyle lim_{xto 0}f(x) = L$
    and $displaystyle lim_{xto 0}g(x)=M$, where $M, N in mathbb R$,
    THEN $displaystyle lim_{xto 0}(f(x)-g(x))=L-M$



    But, since $displaystyle lim_{xto 0} frac{tan x}{x^3} = lim_{xto 0} frac{sin x}{x^3} = infty$, then the theorem does not apply.



    This limit can be evaluated without resorting to L'Hospital.



    begin{align}
    frac{tan x - sin x}{x^3}
    &= frac{frac{sin x}{cos x} - sin x}{x^3} \
    &= frac{sin x - sin x cos x}{x^3 cos x} \
    &= frac{1}{cos x} cdotfrac{sin x}{x} cdot frac{1 - cos x}{x^2} \
    &= frac{1}{cos x} cdotfrac{sin x}{x}
    cdot frac{2sin^2(frac 12x)}{x^2} \
    &= frac{1}{cos x} cdotfrac{sin x}{x}
    cdot frac 12 cdot left(frac{sin frac x2}{frac x2}right)^2 \
    end{align}



    which approaches $dfrac 12$ as $x$ approaches $0$.






    share|cite|improve this answer



















    • 1




      @Steven where did I do $infty - infty$
      – Akash Roy
      Nov 22 at 13:05






    • 1




      Your way is correct.
      – KM101
      Nov 22 at 13:51






    • 1




      @AkashRoy Yes what you did is correct. I was looking at someone elses answer by mistake.
      – steven gregory
      Nov 22 at 14:14








    • 1




      I didn't know we needed to go to Le Hospital to solve limits :)
      – Abraham Zhang
      Nov 22 at 22:36






    • 1




      @stevengregory Check your spelling of "L'Hôpital" and then re-read my comment.
      – Abraham Zhang
      Nov 23 at 1:07
















    28














    This is just another way of saying what the others told you.



    $$lim_{xto 0} frac{tan x - sin x}{x^3}
    ne lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}$$



    The theorem is
    IF $displaystyle lim_{xto 0}f(x) = L$
    and $displaystyle lim_{xto 0}g(x)=M$, where $M, N in mathbb R$,
    THEN $displaystyle lim_{xto 0}(f(x)-g(x))=L-M$



    But, since $displaystyle lim_{xto 0} frac{tan x}{x^3} = lim_{xto 0} frac{sin x}{x^3} = infty$, then the theorem does not apply.



    This limit can be evaluated without resorting to L'Hospital.



    begin{align}
    frac{tan x - sin x}{x^3}
    &= frac{frac{sin x}{cos x} - sin x}{x^3} \
    &= frac{sin x - sin x cos x}{x^3 cos x} \
    &= frac{1}{cos x} cdotfrac{sin x}{x} cdot frac{1 - cos x}{x^2} \
    &= frac{1}{cos x} cdotfrac{sin x}{x}
    cdot frac{2sin^2(frac 12x)}{x^2} \
    &= frac{1}{cos x} cdotfrac{sin x}{x}
    cdot frac 12 cdot left(frac{sin frac x2}{frac x2}right)^2 \
    end{align}



    which approaches $dfrac 12$ as $x$ approaches $0$.






    share|cite|improve this answer



















    • 1




      @Steven where did I do $infty - infty$
      – Akash Roy
      Nov 22 at 13:05






    • 1




      Your way is correct.
      – KM101
      Nov 22 at 13:51






    • 1




      @AkashRoy Yes what you did is correct. I was looking at someone elses answer by mistake.
      – steven gregory
      Nov 22 at 14:14








    • 1




      I didn't know we needed to go to Le Hospital to solve limits :)
      – Abraham Zhang
      Nov 22 at 22:36






    • 1




      @stevengregory Check your spelling of "L'Hôpital" and then re-read my comment.
      – Abraham Zhang
      Nov 23 at 1:07














    28












    28








    28






    This is just another way of saying what the others told you.



    $$lim_{xto 0} frac{tan x - sin x}{x^3}
    ne lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}$$



    The theorem is
    IF $displaystyle lim_{xto 0}f(x) = L$
    and $displaystyle lim_{xto 0}g(x)=M$, where $M, N in mathbb R$,
    THEN $displaystyle lim_{xto 0}(f(x)-g(x))=L-M$



    But, since $displaystyle lim_{xto 0} frac{tan x}{x^3} = lim_{xto 0} frac{sin x}{x^3} = infty$, then the theorem does not apply.



    This limit can be evaluated without resorting to L'Hospital.



    begin{align}
    frac{tan x - sin x}{x^3}
    &= frac{frac{sin x}{cos x} - sin x}{x^3} \
    &= frac{sin x - sin x cos x}{x^3 cos x} \
    &= frac{1}{cos x} cdotfrac{sin x}{x} cdot frac{1 - cos x}{x^2} \
    &= frac{1}{cos x} cdotfrac{sin x}{x}
    cdot frac{2sin^2(frac 12x)}{x^2} \
    &= frac{1}{cos x} cdotfrac{sin x}{x}
    cdot frac 12 cdot left(frac{sin frac x2}{frac x2}right)^2 \
    end{align}



    which approaches $dfrac 12$ as $x$ approaches $0$.






    share|cite|improve this answer














    This is just another way of saying what the others told you.



    $$lim_{xto 0} frac{tan x - sin x}{x^3}
    ne lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}$$



    The theorem is
    IF $displaystyle lim_{xto 0}f(x) = L$
    and $displaystyle lim_{xto 0}g(x)=M$, where $M, N in mathbb R$,
    THEN $displaystyle lim_{xto 0}(f(x)-g(x))=L-M$



    But, since $displaystyle lim_{xto 0} frac{tan x}{x^3} = lim_{xto 0} frac{sin x}{x^3} = infty$, then the theorem does not apply.



    This limit can be evaluated without resorting to L'Hospital.



    begin{align}
    frac{tan x - sin x}{x^3}
    &= frac{frac{sin x}{cos x} - sin x}{x^3} \
    &= frac{sin x - sin x cos x}{x^3 cos x} \
    &= frac{1}{cos x} cdotfrac{sin x}{x} cdot frac{1 - cos x}{x^2} \
    &= frac{1}{cos x} cdotfrac{sin x}{x}
    cdot frac{2sin^2(frac 12x)}{x^2} \
    &= frac{1}{cos x} cdotfrac{sin x}{x}
    cdot frac 12 cdot left(frac{sin frac x2}{frac x2}right)^2 \
    end{align}



    which approaches $dfrac 12$ as $x$ approaches $0$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 21 at 18:19

























    answered Nov 21 at 17:57









    steven gregory

    17.6k32257




    17.6k32257








    • 1




      @Steven where did I do $infty - infty$
      – Akash Roy
      Nov 22 at 13:05






    • 1




      Your way is correct.
      – KM101
      Nov 22 at 13:51






    • 1




      @AkashRoy Yes what you did is correct. I was looking at someone elses answer by mistake.
      – steven gregory
      Nov 22 at 14:14








    • 1




      I didn't know we needed to go to Le Hospital to solve limits :)
      – Abraham Zhang
      Nov 22 at 22:36






    • 1




      @stevengregory Check your spelling of "L'Hôpital" and then re-read my comment.
      – Abraham Zhang
      Nov 23 at 1:07














    • 1




      @Steven where did I do $infty - infty$
      – Akash Roy
      Nov 22 at 13:05






    • 1




      Your way is correct.
      – KM101
      Nov 22 at 13:51






    • 1




      @AkashRoy Yes what you did is correct. I was looking at someone elses answer by mistake.
      – steven gregory
      Nov 22 at 14:14








    • 1




      I didn't know we needed to go to Le Hospital to solve limits :)
      – Abraham Zhang
      Nov 22 at 22:36






    • 1




      @stevengregory Check your spelling of "L'Hôpital" and then re-read my comment.
      – Abraham Zhang
      Nov 23 at 1:07








    1




    1




    @Steven where did I do $infty - infty$
    – Akash Roy
    Nov 22 at 13:05




    @Steven where did I do $infty - infty$
    – Akash Roy
    Nov 22 at 13:05




    1




    1




    Your way is correct.
    – KM101
    Nov 22 at 13:51




    Your way is correct.
    – KM101
    Nov 22 at 13:51




    1




    1




    @AkashRoy Yes what you did is correct. I was looking at someone elses answer by mistake.
    – steven gregory
    Nov 22 at 14:14






    @AkashRoy Yes what you did is correct. I was looking at someone elses answer by mistake.
    – steven gregory
    Nov 22 at 14:14






    1




    1




    I didn't know we needed to go to Le Hospital to solve limits :)
    – Abraham Zhang
    Nov 22 at 22:36




    I didn't know we needed to go to Le Hospital to solve limits :)
    – Abraham Zhang
    Nov 22 at 22:36




    1




    1




    @stevengregory Check your spelling of "L'Hôpital" and then re-read my comment.
    – Abraham Zhang
    Nov 23 at 1:07




    @stevengregory Check your spelling of "L'Hôpital" and then re-read my comment.
    – Abraham Zhang
    Nov 23 at 1:07











    13














    I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ limlimits_{x to 0}big( f(x) - g(x)big)$ is not always equal to $ limlimits_{x to 0} f(x) - limlimits_{x to 0} g(x)$.






    share|cite|improve this answer



















    • 3




      This is the most concise correct answer. The obvious example is to pick any function $f:mathbb Rtomathbb R$ such that $limlimits_{xto 0}$ does not exist. Then $limlimits_{xto 0}big( f(x)-f(x)big)$ exists and equals $0$, but it is not equal to $limlimits_{xto 0}f(x) + limlimits_{xto 0}-f(x)$ since neither one of the latter two limits exists.
      – MPW
      Nov 21 at 18:30










    • There has to be further problems with the work, as splitting works when there's a finite limit. If the OP had worked through the problem correctly, they should have found that the separate limits were indeterminate. The issue is not just splitting, but then recombining them later to cancel out the indeterminate.
      – Acccumulation
      Nov 21 at 19:57










    • Is my answer okay?
      – Akash Roy
      Nov 22 at 8:00
















    13














    I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ limlimits_{x to 0}big( f(x) - g(x)big)$ is not always equal to $ limlimits_{x to 0} f(x) - limlimits_{x to 0} g(x)$.






    share|cite|improve this answer



















    • 3




      This is the most concise correct answer. The obvious example is to pick any function $f:mathbb Rtomathbb R$ such that $limlimits_{xto 0}$ does not exist. Then $limlimits_{xto 0}big( f(x)-f(x)big)$ exists and equals $0$, but it is not equal to $limlimits_{xto 0}f(x) + limlimits_{xto 0}-f(x)$ since neither one of the latter two limits exists.
      – MPW
      Nov 21 at 18:30










    • There has to be further problems with the work, as splitting works when there's a finite limit. If the OP had worked through the problem correctly, they should have found that the separate limits were indeterminate. The issue is not just splitting, but then recombining them later to cancel out the indeterminate.
      – Acccumulation
      Nov 21 at 19:57










    • Is my answer okay?
      – Akash Roy
      Nov 22 at 8:00














    13












    13








    13






    I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ limlimits_{x to 0}big( f(x) - g(x)big)$ is not always equal to $ limlimits_{x to 0} f(x) - limlimits_{x to 0} g(x)$.






    share|cite|improve this answer














    I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ limlimits_{x to 0}big( f(x) - g(x)big)$ is not always equal to $ limlimits_{x to 0} f(x) - limlimits_{x to 0} g(x)$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 21 at 18:26









    MPW

    29.8k11956




    29.8k11956










    answered Nov 21 at 17:55









    Reinstein

    1257




    1257








    • 3




      This is the most concise correct answer. The obvious example is to pick any function $f:mathbb Rtomathbb R$ such that $limlimits_{xto 0}$ does not exist. Then $limlimits_{xto 0}big( f(x)-f(x)big)$ exists and equals $0$, but it is not equal to $limlimits_{xto 0}f(x) + limlimits_{xto 0}-f(x)$ since neither one of the latter two limits exists.
      – MPW
      Nov 21 at 18:30










    • There has to be further problems with the work, as splitting works when there's a finite limit. If the OP had worked through the problem correctly, they should have found that the separate limits were indeterminate. The issue is not just splitting, but then recombining them later to cancel out the indeterminate.
      – Acccumulation
      Nov 21 at 19:57










    • Is my answer okay?
      – Akash Roy
      Nov 22 at 8:00














    • 3




      This is the most concise correct answer. The obvious example is to pick any function $f:mathbb Rtomathbb R$ such that $limlimits_{xto 0}$ does not exist. Then $limlimits_{xto 0}big( f(x)-f(x)big)$ exists and equals $0$, but it is not equal to $limlimits_{xto 0}f(x) + limlimits_{xto 0}-f(x)$ since neither one of the latter two limits exists.
      – MPW
      Nov 21 at 18:30










    • There has to be further problems with the work, as splitting works when there's a finite limit. If the OP had worked through the problem correctly, they should have found that the separate limits were indeterminate. The issue is not just splitting, but then recombining them later to cancel out the indeterminate.
      – Acccumulation
      Nov 21 at 19:57










    • Is my answer okay?
      – Akash Roy
      Nov 22 at 8:00








    3




    3




    This is the most concise correct answer. The obvious example is to pick any function $f:mathbb Rtomathbb R$ such that $limlimits_{xto 0}$ does not exist. Then $limlimits_{xto 0}big( f(x)-f(x)big)$ exists and equals $0$, but it is not equal to $limlimits_{xto 0}f(x) + limlimits_{xto 0}-f(x)$ since neither one of the latter two limits exists.
    – MPW
    Nov 21 at 18:30




    This is the most concise correct answer. The obvious example is to pick any function $f:mathbb Rtomathbb R$ such that $limlimits_{xto 0}$ does not exist. Then $limlimits_{xto 0}big( f(x)-f(x)big)$ exists and equals $0$, but it is not equal to $limlimits_{xto 0}f(x) + limlimits_{xto 0}-f(x)$ since neither one of the latter two limits exists.
    – MPW
    Nov 21 at 18:30












    There has to be further problems with the work, as splitting works when there's a finite limit. If the OP had worked through the problem correctly, they should have found that the separate limits were indeterminate. The issue is not just splitting, but then recombining them later to cancel out the indeterminate.
    – Acccumulation
    Nov 21 at 19:57




    There has to be further problems with the work, as splitting works when there's a finite limit. If the OP had worked through the problem correctly, they should have found that the separate limits were indeterminate. The issue is not just splitting, but then recombining them later to cancel out the indeterminate.
    – Acccumulation
    Nov 21 at 19:57












    Is my answer okay?
    – Akash Roy
    Nov 22 at 8:00




    Is my answer okay?
    – Akash Roy
    Nov 22 at 8:00











    3














    Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.



    We have



    $$lim_{x to 0} tan x= frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . .$$



    $$lim_{x to 0} sin x= frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .$$



    Therefore expression turns to,



    $$lim_{x to 0} frac{frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . . - (frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .)}{x^{3}}$$



    Cancel the $x$ and then enforce the limit after dividing the numerator by $x^{3}$ . The expression simplifies to the calculation of sum of $frac{1}{3}$ and $frac{1}{6}$ which is $color{red} {frac{1}{2}}$






    share|cite|improve this answer























    • Cancel the x/1 term first and then divide the expression by $x^{3}$ then enforce the limit which is x tends to 0 .
      – Akash Roy
      Nov 22 at 5:33












    • Good solution, however I would use truncation (with big-$O$) rather than those dots. And then you don't even need the $x^5$ terms, so you can truncate at $O(x^5)$.
      – Jean-Claude Arbaut
      Nov 22 at 12:51










    • Ok sir Jean Claude Arbaut . Thanks for your words.
      – Akash Roy
      Nov 22 at 13:03










    • That's a nice solution.
      – Raghav
      Nov 22 at 16:24










    • Thanks @Raghav bro.
      – Akash Roy
      Nov 22 at 16:39
















    3














    Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.



    We have



    $$lim_{x to 0} tan x= frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . .$$



    $$lim_{x to 0} sin x= frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .$$



    Therefore expression turns to,



    $$lim_{x to 0} frac{frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . . - (frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .)}{x^{3}}$$



    Cancel the $x$ and then enforce the limit after dividing the numerator by $x^{3}$ . The expression simplifies to the calculation of sum of $frac{1}{3}$ and $frac{1}{6}$ which is $color{red} {frac{1}{2}}$






    share|cite|improve this answer























    • Cancel the x/1 term first and then divide the expression by $x^{3}$ then enforce the limit which is x tends to 0 .
      – Akash Roy
      Nov 22 at 5:33












    • Good solution, however I would use truncation (with big-$O$) rather than those dots. And then you don't even need the $x^5$ terms, so you can truncate at $O(x^5)$.
      – Jean-Claude Arbaut
      Nov 22 at 12:51










    • Ok sir Jean Claude Arbaut . Thanks for your words.
      – Akash Roy
      Nov 22 at 13:03










    • That's a nice solution.
      – Raghav
      Nov 22 at 16:24










    • Thanks @Raghav bro.
      – Akash Roy
      Nov 22 at 16:39














    3












    3








    3






    Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.



    We have



    $$lim_{x to 0} tan x= frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . .$$



    $$lim_{x to 0} sin x= frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .$$



    Therefore expression turns to,



    $$lim_{x to 0} frac{frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . . - (frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .)}{x^{3}}$$



    Cancel the $x$ and then enforce the limit after dividing the numerator by $x^{3}$ . The expression simplifies to the calculation of sum of $frac{1}{3}$ and $frac{1}{6}$ which is $color{red} {frac{1}{2}}$






    share|cite|improve this answer














    Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.



    We have



    $$lim_{x to 0} tan x= frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . .$$



    $$lim_{x to 0} sin x= frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .$$



    Therefore expression turns to,



    $$lim_{x to 0} frac{frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . . - (frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .)}{x^{3}}$$



    Cancel the $x$ and then enforce the limit after dividing the numerator by $x^{3}$ . The expression simplifies to the calculation of sum of $frac{1}{3}$ and $frac{1}{6}$ which is $color{red} {frac{1}{2}}$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 22 at 5:37

























    answered Nov 22 at 5:29









    Akash Roy

    1




    1












    • Cancel the x/1 term first and then divide the expression by $x^{3}$ then enforce the limit which is x tends to 0 .
      – Akash Roy
      Nov 22 at 5:33












    • Good solution, however I would use truncation (with big-$O$) rather than those dots. And then you don't even need the $x^5$ terms, so you can truncate at $O(x^5)$.
      – Jean-Claude Arbaut
      Nov 22 at 12:51










    • Ok sir Jean Claude Arbaut . Thanks for your words.
      – Akash Roy
      Nov 22 at 13:03










    • That's a nice solution.
      – Raghav
      Nov 22 at 16:24










    • Thanks @Raghav bro.
      – Akash Roy
      Nov 22 at 16:39


















    • Cancel the x/1 term first and then divide the expression by $x^{3}$ then enforce the limit which is x tends to 0 .
      – Akash Roy
      Nov 22 at 5:33












    • Good solution, however I would use truncation (with big-$O$) rather than those dots. And then you don't even need the $x^5$ terms, so you can truncate at $O(x^5)$.
      – Jean-Claude Arbaut
      Nov 22 at 12:51










    • Ok sir Jean Claude Arbaut . Thanks for your words.
      – Akash Roy
      Nov 22 at 13:03










    • That's a nice solution.
      – Raghav
      Nov 22 at 16:24










    • Thanks @Raghav bro.
      – Akash Roy
      Nov 22 at 16:39
















    Cancel the x/1 term first and then divide the expression by $x^{3}$ then enforce the limit which is x tends to 0 .
    – Akash Roy
    Nov 22 at 5:33






    Cancel the x/1 term first and then divide the expression by $x^{3}$ then enforce the limit which is x tends to 0 .
    – Akash Roy
    Nov 22 at 5:33














    Good solution, however I would use truncation (with big-$O$) rather than those dots. And then you don't even need the $x^5$ terms, so you can truncate at $O(x^5)$.
    – Jean-Claude Arbaut
    Nov 22 at 12:51




    Good solution, however I would use truncation (with big-$O$) rather than those dots. And then you don't even need the $x^5$ terms, so you can truncate at $O(x^5)$.
    – Jean-Claude Arbaut
    Nov 22 at 12:51












    Ok sir Jean Claude Arbaut . Thanks for your words.
    – Akash Roy
    Nov 22 at 13:03




    Ok sir Jean Claude Arbaut . Thanks for your words.
    – Akash Roy
    Nov 22 at 13:03












    That's a nice solution.
    – Raghav
    Nov 22 at 16:24




    That's a nice solution.
    – Raghav
    Nov 22 at 16:24












    Thanks @Raghav bro.
    – Akash Roy
    Nov 22 at 16:39




    Thanks @Raghav bro.
    – Akash Roy
    Nov 22 at 16:39











    2














    Don't try to take the limit of each seperately & then take the difference - you just get the difference between two infinities! Express $sin$ & $tan$ as Taylor series - each has first term in $theta^1$ with coefficient 1, so in the difference it drops out. If you plot $sintheta-tantheta$ it looks like a cubic at the origin. Then if you divide that series by $theta^3$, & you get a series with an initial term in $theta^0$, ie a constant term. (This is shown in plots: if you plot that curve just described, ÷by $theta^3$, it begins somewhere along the y -axis instead of at the origin.) This is then all that is left as $thetarightarrow 0$. That's equivalent to tracing the plot I have just described in parenthesis to its point of intersection with the y -axis.



    To actually get the answer immediately you just subtract the coefficient for $theta^3$ in the series for $sintheta$ from that in that for $tantheta$, & you get 1/3 - -1/6 = 1/2.






    share|cite|improve this answer




























      2














      Don't try to take the limit of each seperately & then take the difference - you just get the difference between two infinities! Express $sin$ & $tan$ as Taylor series - each has first term in $theta^1$ with coefficient 1, so in the difference it drops out. If you plot $sintheta-tantheta$ it looks like a cubic at the origin. Then if you divide that series by $theta^3$, & you get a series with an initial term in $theta^0$, ie a constant term. (This is shown in plots: if you plot that curve just described, ÷by $theta^3$, it begins somewhere along the y -axis instead of at the origin.) This is then all that is left as $thetarightarrow 0$. That's equivalent to tracing the plot I have just described in parenthesis to its point of intersection with the y -axis.



      To actually get the answer immediately you just subtract the coefficient for $theta^3$ in the series for $sintheta$ from that in that for $tantheta$, & you get 1/3 - -1/6 = 1/2.






      share|cite|improve this answer


























        2












        2








        2






        Don't try to take the limit of each seperately & then take the difference - you just get the difference between two infinities! Express $sin$ & $tan$ as Taylor series - each has first term in $theta^1$ with coefficient 1, so in the difference it drops out. If you plot $sintheta-tantheta$ it looks like a cubic at the origin. Then if you divide that series by $theta^3$, & you get a series with an initial term in $theta^0$, ie a constant term. (This is shown in plots: if you plot that curve just described, ÷by $theta^3$, it begins somewhere along the y -axis instead of at the origin.) This is then all that is left as $thetarightarrow 0$. That's equivalent to tracing the plot I have just described in parenthesis to its point of intersection with the y -axis.



        To actually get the answer immediately you just subtract the coefficient for $theta^3$ in the series for $sintheta$ from that in that for $tantheta$, & you get 1/3 - -1/6 = 1/2.






        share|cite|improve this answer














        Don't try to take the limit of each seperately & then take the difference - you just get the difference between two infinities! Express $sin$ & $tan$ as Taylor series - each has first term in $theta^1$ with coefficient 1, so in the difference it drops out. If you plot $sintheta-tantheta$ it looks like a cubic at the origin. Then if you divide that series by $theta^3$, & you get a series with an initial term in $theta^0$, ie a constant term. (This is shown in plots: if you plot that curve just described, ÷by $theta^3$, it begins somewhere along the y -axis instead of at the origin.) This is then all that is left as $thetarightarrow 0$. That's equivalent to tracing the plot I have just described in parenthesis to its point of intersection with the y -axis.



        To actually get the answer immediately you just subtract the coefficient for $theta^3$ in the series for $sintheta$ from that in that for $tantheta$, & you get 1/3 - -1/6 = 1/2.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 22 at 4:39

























        answered Nov 22 at 4:26









        AmbretteOrrisey

        56710




        56710






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008071%2flim-limits-x-to-0-frac-tan-x-sin-xx3%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Aardman Animations

            Are they similar matrix

            “minimization” problem in Euclidean space related to orthonormal basis