The idea of having a limit as $x rightarrow a$ where the denominator is 0 confuses me












0












$begingroup$


What does that mean? Say as $x rightarrow 2$, our equation looks like this, $2x + 1 over {x - 2}$.



Then the denominator will go towards 0. Now how would I find the limit of this? How do I know it will exist?



Also, can someone help me know why the limit of $e^x over {e^x + 1}$ as $x rightarrow -infty$ is $0$? and when it's approaching $+ infty$ the limit is 1? And why that represents the horizontal asymptote?






calculus







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    What does that mean? Say as $x rightarrow 2$, our equation looks like this, $2x + 1 over {x - 2}$.



    Then the denominator will go towards 0. Now how would I find the limit of this? How do I know it will exist?



    Also, can someone help me know why the limit of $e^x over {e^x + 1}$ as $x rightarrow -infty$ is $0$? and when it's approaching $+ infty$ the limit is 1? And why that represents the horizontal asymptote?






    calculus







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      What does that mean? Say as $x rightarrow 2$, our equation looks like this, $2x + 1 over {x - 2}$.



      Then the denominator will go towards 0. Now how would I find the limit of this? How do I know it will exist?



      Also, can someone help me know why the limit of $e^x over {e^x + 1}$ as $x rightarrow -infty$ is $0$? and when it's approaching $+ infty$ the limit is 1? And why that represents the horizontal asymptote?






      calculus







      share|cite|improve this question









      $endgroup$




      What does that mean? Say as $x rightarrow 2$, our equation looks like this, $2x + 1 over {x - 2}$.



      Then the denominator will go towards 0. Now how would I find the limit of this? How do I know it will exist?



      Also, can someone help me know why the limit of $e^x over {e^x + 1}$ as $x rightarrow -infty$ is $0$? and when it's approaching $+ infty$ the limit is 1? And why that represents the horizontal asymptote?






      calculus




      calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 10 '18 at 5:53









      mingming

      3415




      3415






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          First off, $dfrac{2x+1}{x-2}$ is not an "equation" since there is no equal sign. You probably want to call that a mathematical expression or a function (and then you "take the limit" of the function as $xrightarrow 2$).



          In order for the "bilateral" limit as $xrightarrow 2$ to exist, the "unilateral" limits (left and right) must exist and be equal.



          Replacing $x$ by $2^-$ yields $2^--2=0^-$ in which case your function takes the form $dfrac{text{+}}{0^-}rightarrow-infty$



          Replacing $x$ by $2^+$ yields $2^+-2=0^+$ in which case your function takes the form $dfrac{text{+}}{0^+}rightarrow+infty$



          The unilateral limits are not equal (I like to say they "don't agree") so the limit as $xrightarrow2$ does not exist.



          Another suggestion for the second part:



          $$f(x)=frac{e^x}{e^x+1}=frac{e^x}{e^x(1+e^{-x})}=frac{1}{1+e^{-x}}$$



          If $xrightarrow+infty$, then $f(x)rightarrowfrac{1}{1+e^{-infty}}=frac{1}{1+0}=1$



          If $xrightarrow-infty$, then $f(x)rightarrowfrac{1}{1+e^{-(-infty))}}=frac{1}{1+infty}=0$



          To clarify, if you have $frac{0}{0}$ or $pmfrac{infty}{infty}$, it's indeterminate and factoring might help. Otherwise:



          $$frac{+}{0^+}rightarrow+infty,quadfrac{-}{0^+}rightarrow-infty,quadfrac{+}{0^-}rightarrow-infty,quadfrac{-}{0^-}rightarrow+infty,$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So, those are the horizontal limits correct? For the bottom example? And so when we plug in infinity into $e^x$ we just simulate what will happen to those specific functions?
            $endgroup$
            – ming
            Dec 10 '18 at 6:31










          • $begingroup$
            Also can you let me know if I have this rule understood correctly? We only use the unilateral and bilateral limit rule if x is approaching a certain point to find what the limit is @ point "a"? Otherwise we don't use that rule?
            $endgroup$
            – ming
            Dec 10 '18 at 6:34










          • $begingroup$
            @ming More specifically, these last 2 limits indicate the behavior of the function near $+infty$ and $-infty$ respectively and here. Here, the behavior is asymptotic so the function as horizontal asymptotes ($y=0$ and $y=1$). The last part where I plug-in $infty$ might be a bit sketchy but I find it helpful to understand why the limits both equal $1$ and $0$.
            $endgroup$
            – orion2112
            Dec 10 '18 at 6:36










          • $begingroup$
            @ming You can't use the left/right limit when $xrightarrowpminfty$ (because you're approaching infinity from only one possible direction, you can't approach it "from further"). What I meant by "bilateral" limit is just the limit itself (approaching the value from both left and right at the same time). This exists if the unilateral limits exist and are equal. We mainly "check" for the unilateral limits if we see a division by $0$ or for functions defined by parts.
            $endgroup$
            – orion2112
            Dec 10 '18 at 6:45










          • $begingroup$
            Cool! So only use both sided limits if we see division by 0 or a piecewise function.
            $endgroup$
            – ming
            Dec 10 '18 at 6:48



















          0












          $begingroup$

          For $frac{2x+1}{x-2}$, it may help to sketch a plot by yourself. When $x$ approaches $2$ from above, $frac{2x+1}{x-2}$ gets arbitrarily large.
          When $x$ approaches $2$ from below, $frac{2x+1}{x-2}$ gets arbitrarily small (negative). So, the limit does not exist.




          For $frac{e^x}{e^x+1}$, if $x to -infty$, then $e^x to 0$, so
          $frac{e^x}{e^x+1} to frac{0}{0+1}$.



          When taking $x to infty$, it may help to write $frac{e^x}{e^x+1} = 1 - frac{1}{e^x+1}$.






          share|cite|improve this answer









          $endgroup$













            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%2f3033514%2fthe-idea-of-having-a-limit-as-x-rightarrow-a-where-the-denominator-is-0-confu%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            First off, $dfrac{2x+1}{x-2}$ is not an "equation" since there is no equal sign. You probably want to call that a mathematical expression or a function (and then you "take the limit" of the function as $xrightarrow 2$).



            In order for the "bilateral" limit as $xrightarrow 2$ to exist, the "unilateral" limits (left and right) must exist and be equal.



            Replacing $x$ by $2^-$ yields $2^--2=0^-$ in which case your function takes the form $dfrac{text{+}}{0^-}rightarrow-infty$



            Replacing $x$ by $2^+$ yields $2^+-2=0^+$ in which case your function takes the form $dfrac{text{+}}{0^+}rightarrow+infty$



            The unilateral limits are not equal (I like to say they "don't agree") so the limit as $xrightarrow2$ does not exist.



            Another suggestion for the second part:



            $$f(x)=frac{e^x}{e^x+1}=frac{e^x}{e^x(1+e^{-x})}=frac{1}{1+e^{-x}}$$



            If $xrightarrow+infty$, then $f(x)rightarrowfrac{1}{1+e^{-infty}}=frac{1}{1+0}=1$



            If $xrightarrow-infty$, then $f(x)rightarrowfrac{1}{1+e^{-(-infty))}}=frac{1}{1+infty}=0$



            To clarify, if you have $frac{0}{0}$ or $pmfrac{infty}{infty}$, it's indeterminate and factoring might help. Otherwise:



            $$frac{+}{0^+}rightarrow+infty,quadfrac{-}{0^+}rightarrow-infty,quadfrac{+}{0^-}rightarrow-infty,quadfrac{-}{0^-}rightarrow+infty,$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              So, those are the horizontal limits correct? For the bottom example? And so when we plug in infinity into $e^x$ we just simulate what will happen to those specific functions?
              $endgroup$
              – ming
              Dec 10 '18 at 6:31










            • $begingroup$
              Also can you let me know if I have this rule understood correctly? We only use the unilateral and bilateral limit rule if x is approaching a certain point to find what the limit is @ point "a"? Otherwise we don't use that rule?
              $endgroup$
              – ming
              Dec 10 '18 at 6:34










            • $begingroup$
              @ming More specifically, these last 2 limits indicate the behavior of the function near $+infty$ and $-infty$ respectively and here. Here, the behavior is asymptotic so the function as horizontal asymptotes ($y=0$ and $y=1$). The last part where I plug-in $infty$ might be a bit sketchy but I find it helpful to understand why the limits both equal $1$ and $0$.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:36










            • $begingroup$
              @ming You can't use the left/right limit when $xrightarrowpminfty$ (because you're approaching infinity from only one possible direction, you can't approach it "from further"). What I meant by "bilateral" limit is just the limit itself (approaching the value from both left and right at the same time). This exists if the unilateral limits exist and are equal. We mainly "check" for the unilateral limits if we see a division by $0$ or for functions defined by parts.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:45










            • $begingroup$
              Cool! So only use both sided limits if we see division by 0 or a piecewise function.
              $endgroup$
              – ming
              Dec 10 '18 at 6:48
















            1












            $begingroup$

            First off, $dfrac{2x+1}{x-2}$ is not an "equation" since there is no equal sign. You probably want to call that a mathematical expression or a function (and then you "take the limit" of the function as $xrightarrow 2$).



            In order for the "bilateral" limit as $xrightarrow 2$ to exist, the "unilateral" limits (left and right) must exist and be equal.



            Replacing $x$ by $2^-$ yields $2^--2=0^-$ in which case your function takes the form $dfrac{text{+}}{0^-}rightarrow-infty$



            Replacing $x$ by $2^+$ yields $2^+-2=0^+$ in which case your function takes the form $dfrac{text{+}}{0^+}rightarrow+infty$



            The unilateral limits are not equal (I like to say they "don't agree") so the limit as $xrightarrow2$ does not exist.



            Another suggestion for the second part:



            $$f(x)=frac{e^x}{e^x+1}=frac{e^x}{e^x(1+e^{-x})}=frac{1}{1+e^{-x}}$$



            If $xrightarrow+infty$, then $f(x)rightarrowfrac{1}{1+e^{-infty}}=frac{1}{1+0}=1$



            If $xrightarrow-infty$, then $f(x)rightarrowfrac{1}{1+e^{-(-infty))}}=frac{1}{1+infty}=0$



            To clarify, if you have $frac{0}{0}$ or $pmfrac{infty}{infty}$, it's indeterminate and factoring might help. Otherwise:



            $$frac{+}{0^+}rightarrow+infty,quadfrac{-}{0^+}rightarrow-infty,quadfrac{+}{0^-}rightarrow-infty,quadfrac{-}{0^-}rightarrow+infty,$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              So, those are the horizontal limits correct? For the bottom example? And so when we plug in infinity into $e^x$ we just simulate what will happen to those specific functions?
              $endgroup$
              – ming
              Dec 10 '18 at 6:31










            • $begingroup$
              Also can you let me know if I have this rule understood correctly? We only use the unilateral and bilateral limit rule if x is approaching a certain point to find what the limit is @ point "a"? Otherwise we don't use that rule?
              $endgroup$
              – ming
              Dec 10 '18 at 6:34










            • $begingroup$
              @ming More specifically, these last 2 limits indicate the behavior of the function near $+infty$ and $-infty$ respectively and here. Here, the behavior is asymptotic so the function as horizontal asymptotes ($y=0$ and $y=1$). The last part where I plug-in $infty$ might be a bit sketchy but I find it helpful to understand why the limits both equal $1$ and $0$.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:36










            • $begingroup$
              @ming You can't use the left/right limit when $xrightarrowpminfty$ (because you're approaching infinity from only one possible direction, you can't approach it "from further"). What I meant by "bilateral" limit is just the limit itself (approaching the value from both left and right at the same time). This exists if the unilateral limits exist and are equal. We mainly "check" for the unilateral limits if we see a division by $0$ or for functions defined by parts.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:45










            • $begingroup$
              Cool! So only use both sided limits if we see division by 0 or a piecewise function.
              $endgroup$
              – ming
              Dec 10 '18 at 6:48














            1












            1








            1





            $begingroup$

            First off, $dfrac{2x+1}{x-2}$ is not an "equation" since there is no equal sign. You probably want to call that a mathematical expression or a function (and then you "take the limit" of the function as $xrightarrow 2$).



            In order for the "bilateral" limit as $xrightarrow 2$ to exist, the "unilateral" limits (left and right) must exist and be equal.



            Replacing $x$ by $2^-$ yields $2^--2=0^-$ in which case your function takes the form $dfrac{text{+}}{0^-}rightarrow-infty$



            Replacing $x$ by $2^+$ yields $2^+-2=0^+$ in which case your function takes the form $dfrac{text{+}}{0^+}rightarrow+infty$



            The unilateral limits are not equal (I like to say they "don't agree") so the limit as $xrightarrow2$ does not exist.



            Another suggestion for the second part:



            $$f(x)=frac{e^x}{e^x+1}=frac{e^x}{e^x(1+e^{-x})}=frac{1}{1+e^{-x}}$$



            If $xrightarrow+infty$, then $f(x)rightarrowfrac{1}{1+e^{-infty}}=frac{1}{1+0}=1$



            If $xrightarrow-infty$, then $f(x)rightarrowfrac{1}{1+e^{-(-infty))}}=frac{1}{1+infty}=0$



            To clarify, if you have $frac{0}{0}$ or $pmfrac{infty}{infty}$, it's indeterminate and factoring might help. Otherwise:



            $$frac{+}{0^+}rightarrow+infty,quadfrac{-}{0^+}rightarrow-infty,quadfrac{+}{0^-}rightarrow-infty,quadfrac{-}{0^-}rightarrow+infty,$$






            share|cite|improve this answer











            $endgroup$



            First off, $dfrac{2x+1}{x-2}$ is not an "equation" since there is no equal sign. You probably want to call that a mathematical expression or a function (and then you "take the limit" of the function as $xrightarrow 2$).



            In order for the "bilateral" limit as $xrightarrow 2$ to exist, the "unilateral" limits (left and right) must exist and be equal.



            Replacing $x$ by $2^-$ yields $2^--2=0^-$ in which case your function takes the form $dfrac{text{+}}{0^-}rightarrow-infty$



            Replacing $x$ by $2^+$ yields $2^+-2=0^+$ in which case your function takes the form $dfrac{text{+}}{0^+}rightarrow+infty$



            The unilateral limits are not equal (I like to say they "don't agree") so the limit as $xrightarrow2$ does not exist.



            Another suggestion for the second part:



            $$f(x)=frac{e^x}{e^x+1}=frac{e^x}{e^x(1+e^{-x})}=frac{1}{1+e^{-x}}$$



            If $xrightarrow+infty$, then $f(x)rightarrowfrac{1}{1+e^{-infty}}=frac{1}{1+0}=1$



            If $xrightarrow-infty$, then $f(x)rightarrowfrac{1}{1+e^{-(-infty))}}=frac{1}{1+infty}=0$



            To clarify, if you have $frac{0}{0}$ or $pmfrac{infty}{infty}$, it's indeterminate and factoring might help. Otherwise:



            $$frac{+}{0^+}rightarrow+infty,quadfrac{-}{0^+}rightarrow-infty,quadfrac{+}{0^-}rightarrow-infty,quadfrac{-}{0^-}rightarrow+infty,$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 10 '18 at 7:06

























            answered Dec 10 '18 at 6:21









            orion2112orion2112

            461310




            461310












            • $begingroup$
              So, those are the horizontal limits correct? For the bottom example? And so when we plug in infinity into $e^x$ we just simulate what will happen to those specific functions?
              $endgroup$
              – ming
              Dec 10 '18 at 6:31










            • $begingroup$
              Also can you let me know if I have this rule understood correctly? We only use the unilateral and bilateral limit rule if x is approaching a certain point to find what the limit is @ point "a"? Otherwise we don't use that rule?
              $endgroup$
              – ming
              Dec 10 '18 at 6:34










            • $begingroup$
              @ming More specifically, these last 2 limits indicate the behavior of the function near $+infty$ and $-infty$ respectively and here. Here, the behavior is asymptotic so the function as horizontal asymptotes ($y=0$ and $y=1$). The last part where I plug-in $infty$ might be a bit sketchy but I find it helpful to understand why the limits both equal $1$ and $0$.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:36










            • $begingroup$
              @ming You can't use the left/right limit when $xrightarrowpminfty$ (because you're approaching infinity from only one possible direction, you can't approach it "from further"). What I meant by "bilateral" limit is just the limit itself (approaching the value from both left and right at the same time). This exists if the unilateral limits exist and are equal. We mainly "check" for the unilateral limits if we see a division by $0$ or for functions defined by parts.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:45










            • $begingroup$
              Cool! So only use both sided limits if we see division by 0 or a piecewise function.
              $endgroup$
              – ming
              Dec 10 '18 at 6:48


















            • $begingroup$
              So, those are the horizontal limits correct? For the bottom example? And so when we plug in infinity into $e^x$ we just simulate what will happen to those specific functions?
              $endgroup$
              – ming
              Dec 10 '18 at 6:31










            • $begingroup$
              Also can you let me know if I have this rule understood correctly? We only use the unilateral and bilateral limit rule if x is approaching a certain point to find what the limit is @ point "a"? Otherwise we don't use that rule?
              $endgroup$
              – ming
              Dec 10 '18 at 6:34










            • $begingroup$
              @ming More specifically, these last 2 limits indicate the behavior of the function near $+infty$ and $-infty$ respectively and here. Here, the behavior is asymptotic so the function as horizontal asymptotes ($y=0$ and $y=1$). The last part where I plug-in $infty$ might be a bit sketchy but I find it helpful to understand why the limits both equal $1$ and $0$.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:36










            • $begingroup$
              @ming You can't use the left/right limit when $xrightarrowpminfty$ (because you're approaching infinity from only one possible direction, you can't approach it "from further"). What I meant by "bilateral" limit is just the limit itself (approaching the value from both left and right at the same time). This exists if the unilateral limits exist and are equal. We mainly "check" for the unilateral limits if we see a division by $0$ or for functions defined by parts.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:45










            • $begingroup$
              Cool! So only use both sided limits if we see division by 0 or a piecewise function.
              $endgroup$
              – ming
              Dec 10 '18 at 6:48
















            $begingroup$
            So, those are the horizontal limits correct? For the bottom example? And so when we plug in infinity into $e^x$ we just simulate what will happen to those specific functions?
            $endgroup$
            – ming
            Dec 10 '18 at 6:31




            $begingroup$
            So, those are the horizontal limits correct? For the bottom example? And so when we plug in infinity into $e^x$ we just simulate what will happen to those specific functions?
            $endgroup$
            – ming
            Dec 10 '18 at 6:31












            $begingroup$
            Also can you let me know if I have this rule understood correctly? We only use the unilateral and bilateral limit rule if x is approaching a certain point to find what the limit is @ point "a"? Otherwise we don't use that rule?
            $endgroup$
            – ming
            Dec 10 '18 at 6:34




            $begingroup$
            Also can you let me know if I have this rule understood correctly? We only use the unilateral and bilateral limit rule if x is approaching a certain point to find what the limit is @ point "a"? Otherwise we don't use that rule?
            $endgroup$
            – ming
            Dec 10 '18 at 6:34












            $begingroup$
            @ming More specifically, these last 2 limits indicate the behavior of the function near $+infty$ and $-infty$ respectively and here. Here, the behavior is asymptotic so the function as horizontal asymptotes ($y=0$ and $y=1$). The last part where I plug-in $infty$ might be a bit sketchy but I find it helpful to understand why the limits both equal $1$ and $0$.
            $endgroup$
            – orion2112
            Dec 10 '18 at 6:36




            $begingroup$
            @ming More specifically, these last 2 limits indicate the behavior of the function near $+infty$ and $-infty$ respectively and here. Here, the behavior is asymptotic so the function as horizontal asymptotes ($y=0$ and $y=1$). The last part where I plug-in $infty$ might be a bit sketchy but I find it helpful to understand why the limits both equal $1$ and $0$.
            $endgroup$
            – orion2112
            Dec 10 '18 at 6:36












            $begingroup$
            @ming You can't use the left/right limit when $xrightarrowpminfty$ (because you're approaching infinity from only one possible direction, you can't approach it "from further"). What I meant by "bilateral" limit is just the limit itself (approaching the value from both left and right at the same time). This exists if the unilateral limits exist and are equal. We mainly "check" for the unilateral limits if we see a division by $0$ or for functions defined by parts.
            $endgroup$
            – orion2112
            Dec 10 '18 at 6:45




            $begingroup$
            @ming You can't use the left/right limit when $xrightarrowpminfty$ (because you're approaching infinity from only one possible direction, you can't approach it "from further"). What I meant by "bilateral" limit is just the limit itself (approaching the value from both left and right at the same time). This exists if the unilateral limits exist and are equal. We mainly "check" for the unilateral limits if we see a division by $0$ or for functions defined by parts.
            $endgroup$
            – orion2112
            Dec 10 '18 at 6:45












            $begingroup$
            Cool! So only use both sided limits if we see division by 0 or a piecewise function.
            $endgroup$
            – ming
            Dec 10 '18 at 6:48




            $begingroup$
            Cool! So only use both sided limits if we see division by 0 or a piecewise function.
            $endgroup$
            – ming
            Dec 10 '18 at 6:48











            0












            $begingroup$

            For $frac{2x+1}{x-2}$, it may help to sketch a plot by yourself. When $x$ approaches $2$ from above, $frac{2x+1}{x-2}$ gets arbitrarily large.
            When $x$ approaches $2$ from below, $frac{2x+1}{x-2}$ gets arbitrarily small (negative). So, the limit does not exist.




            For $frac{e^x}{e^x+1}$, if $x to -infty$, then $e^x to 0$, so
            $frac{e^x}{e^x+1} to frac{0}{0+1}$.



            When taking $x to infty$, it may help to write $frac{e^x}{e^x+1} = 1 - frac{1}{e^x+1}$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              For $frac{2x+1}{x-2}$, it may help to sketch a plot by yourself. When $x$ approaches $2$ from above, $frac{2x+1}{x-2}$ gets arbitrarily large.
              When $x$ approaches $2$ from below, $frac{2x+1}{x-2}$ gets arbitrarily small (negative). So, the limit does not exist.




              For $frac{e^x}{e^x+1}$, if $x to -infty$, then $e^x to 0$, so
              $frac{e^x}{e^x+1} to frac{0}{0+1}$.



              When taking $x to infty$, it may help to write $frac{e^x}{e^x+1} = 1 - frac{1}{e^x+1}$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                For $frac{2x+1}{x-2}$, it may help to sketch a plot by yourself. When $x$ approaches $2$ from above, $frac{2x+1}{x-2}$ gets arbitrarily large.
                When $x$ approaches $2$ from below, $frac{2x+1}{x-2}$ gets arbitrarily small (negative). So, the limit does not exist.




                For $frac{e^x}{e^x+1}$, if $x to -infty$, then $e^x to 0$, so
                $frac{e^x}{e^x+1} to frac{0}{0+1}$.



                When taking $x to infty$, it may help to write $frac{e^x}{e^x+1} = 1 - frac{1}{e^x+1}$.






                share|cite|improve this answer









                $endgroup$



                For $frac{2x+1}{x-2}$, it may help to sketch a plot by yourself. When $x$ approaches $2$ from above, $frac{2x+1}{x-2}$ gets arbitrarily large.
                When $x$ approaches $2$ from below, $frac{2x+1}{x-2}$ gets arbitrarily small (negative). So, the limit does not exist.




                For $frac{e^x}{e^x+1}$, if $x to -infty$, then $e^x to 0$, so
                $frac{e^x}{e^x+1} to frac{0}{0+1}$.



                When taking $x to infty$, it may help to write $frac{e^x}{e^x+1} = 1 - frac{1}{e^x+1}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 10 '18 at 6:01









                angryavianangryavian

                40.7k23380




                40.7k23380






























                    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.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033514%2fthe-idea-of-having-a-limit-as-x-rightarrow-a-where-the-denominator-is-0-confu%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

                    Probability when a professor distributes a quiz and homework assignment to a class of n students.

                    Image
                    0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                    Read more

                    Aardman Animations

                    Image
                    Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                    Read more

                    Are they similar matrix

                    Image
                    0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                    Read more