How to evaluate $lim_{xto 0} frac {(sin(2x)-2sin(x))^4}{(3+cos(2x)-4cos(x))^3}$?
$begingroup$
$$lim_{xto 0} frac {(sin(2x)-2sin(x))^4}{(3+cos(2x)-4cos(x))^3}$$
without L'Hôpital.
I've tried using equivalences with ${(sin(2x)-2sin(x))^4}$ and arrived at $-x^{12}$ but I don't know how to handle ${(3+cos(2x)-4cos(x))^3}$. Using $cos(2x)=cos^2(x)-sin^2(x)$ hasn't helped, so any hint?
real-analysis limits-without-lhopital
edited Jan 8 at 18:17
Andrei
13.8k21230
asked Jan 8 at 18:14
iggykimiiggykimi
31910
$endgroup$
add a comment |
$begingroup$
$$lim_{xto 0} frac {(sin(2x)-2sin(x))^4}{(3+cos(2x)-4cos(x))^3}$$
without L'Hôpital.
I've tried using equivalences with ${(sin(2x)-2sin(x))^4}$ and arrived at $-x^{12}$ but I don't know how to handle ${(3+cos(2x)-4cos(x))^3}$. Using $cos(2x)=cos^2(x)-sin^2(x)$ hasn't helped, so any hint?
real-analysis limits-without-lhopital
edited Jan 8 at 18:17
Andrei
13.8k21230
asked Jan 8 at 18:14
iggykimiiggykimi
31910
$endgroup$
add a comment |
$begingroup$
$$lim_{xto 0} frac {(sin(2x)-2sin(x))^4}{(3+cos(2x)-4cos(x))^3}$$
without L'Hôpital.
I've tried using equivalences with ${(sin(2x)-2sin(x))^4}$ and arrived at $-x^{12}$ but I don't know how to handle ${(3+cos(2x)-4cos(x))^3}$. Using $cos(2x)=cos^2(x)-sin^2(x)$ hasn't helped, so any hint?
real-analysis limits-without-lhopital
edited Jan 8 at 18:17
Andrei
13.8k21230
asked Jan 8 at 18:14
iggykimiiggykimi
31910
$endgroup$
$$lim_{xto 0} frac {(sin(2x)-2sin(x))^4}{(3+cos(2x)-4cos(x))^3}$$
without L'Hôpital.
I've tried using equivalences with ${(sin(2x)-2sin(x))^4}$ and arrived at $-x^{12}$ but I don't know how to handle ${(3+cos(2x)-4cos(x))^3}$. Using $cos(2x)=cos^2(x)-sin^2(x)$ hasn't helped, so any hint?
real-analysis limits-without-lhopital
real-analysis limits-without-lhopital
edited Jan 8 at 18:17
Andrei
13.8k21230
asked Jan 8 at 18:14
iggykimiiggykimi
31910
edited Jan 8 at 18:17
Andrei
13.8k21230
asked Jan 8 at 18:14
iggykimiiggykimi
31910
edited Jan 8 at 18:17
Andrei
13.8k21230
edited Jan 8 at 18:17
Andrei
13.8k21230
edited Jan 8 at 18:17
Andrei
13.8k21230
13.8k21230
asked Jan 8 at 18:14
iggykimiiggykimi
31910
asked Jan 8 at 18:14
iggykimiiggykimi
31910
asked Jan 8 at 18:14
iggykimiiggykimi
31910
31910
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: Note that
$$ 3+cos(2x)-4cos(x) = 3 + 2cos^2(x) - 1 - 4cos(x) = 2(cos(x)-1)^2, $$
and that
$$ sin(2x) - 2sin(x) = 2sin(x)cos(x)-2sin(x) = 2sin(x)(cos(x)-1). $$
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8951625
$endgroup$
add a comment |
$begingroup$
Hint: Your quotient can be simplified to $$8cosleft(frac{x}{2}right)^4$$
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
add a comment |
$begingroup$
One can evaluate all such lims using series expansions.
$$
sin(x) = x - frac{x^3}{6} + o(x^4)\
cos(x) = 1 - frac{x^2}{2} + o(x^4)
$$
https://en.wikipedia.org/wiki/Taylor_series#Trigonometric_functions
Just substitute functions with their expansions. Then just find lim of expression~polynomial/polynomial. Add more $x^n$ terms, if first 2 is not enough.
L'Hôpital's rule is another form of this more general approach.
answered Jan 8 at 18:30
Mike_Mike_
616
$endgroup$
add a comment |
Your Answer
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066530%2fhow-to-evaluate-lim-x-to-0-frac-sin2x-2-sinx43-cos2x-4-cosx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Note that
$$ 3+cos(2x)-4cos(x) = 3 + 2cos^2(x) - 1 - 4cos(x) = 2(cos(x)-1)^2, $$
and that
$$ sin(2x) - 2sin(x) = 2sin(x)cos(x)-2sin(x) = 2sin(x)(cos(x)-1). $$
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8951625
$endgroup$
add a comment |
$begingroup$
Hint: Note that
$$ 3+cos(2x)-4cos(x) = 3 + 2cos^2(x) - 1 - 4cos(x) = 2(cos(x)-1)^2, $$
and that
$$ sin(2x) - 2sin(x) = 2sin(x)cos(x)-2sin(x) = 2sin(x)(cos(x)-1). $$
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8951625
$endgroup$
add a comment |
$begingroup$
Hint: Note that
$$ 3+cos(2x)-4cos(x) = 3 + 2cos^2(x) - 1 - 4cos(x) = 2(cos(x)-1)^2, $$
and that
$$ sin(2x) - 2sin(x) = 2sin(x)cos(x)-2sin(x) = 2sin(x)(cos(x)-1). $$
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8951625
$endgroup$
Hint: Note that
$$ 3+cos(2x)-4cos(x) = 3 + 2cos^2(x) - 1 - 4cos(x) = 2(cos(x)-1)^2, $$
and that
$$ sin(2x) - 2sin(x) = 2sin(x)cos(x)-2sin(x) = 2sin(x)(cos(x)-1). $$
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8951625
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8951625
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8951625
answered Jan 8 at 18:17
MisterRiemannMisterRiemann
5,8951625
5,8951625
add a comment |
add a comment |
$begingroup$
Hint: Your quotient can be simplified to $$8cosleft(frac{x}{2}right)^4$$
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
add a comment |
$begingroup$
Hint: Your quotient can be simplified to $$8cosleft(frac{x}{2}right)^4$$
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
add a comment |
$begingroup$
Hint: Your quotient can be simplified to $$8cosleft(frac{x}{2}right)^4$$
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
$endgroup$
Hint: Your quotient can be simplified to $$8cosleft(frac{x}{2}right)^4$$
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
answered Jan 8 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79k42867
79k42867
add a comment |
add a comment |
$begingroup$
One can evaluate all such lims using series expansions.
$$
sin(x) = x - frac{x^3}{6} + o(x^4)\
cos(x) = 1 - frac{x^2}{2} + o(x^4)
$$
https://en.wikipedia.org/wiki/Taylor_series#Trigonometric_functions
Just substitute functions with their expansions. Then just find lim of expression~polynomial/polynomial. Add more $x^n$ terms, if first 2 is not enough.
L'Hôpital's rule is another form of this more general approach.
answered Jan 8 at 18:30
Mike_Mike_
616
$endgroup$
add a comment |
$begingroup$
One can evaluate all such lims using series expansions.
$$
sin(x) = x - frac{x^3}{6} + o(x^4)\
cos(x) = 1 - frac{x^2}{2} + o(x^4)
$$
https://en.wikipedia.org/wiki/Taylor_series#Trigonometric_functions
Just substitute functions with their expansions. Then just find lim of expression~polynomial/polynomial. Add more $x^n$ terms, if first 2 is not enough.
L'Hôpital's rule is another form of this more general approach.
answered Jan 8 at 18:30
Mike_Mike_
616
$endgroup$
add a comment |
$begingroup$
One can evaluate all such lims using series expansions.
$$
sin(x) = x - frac{x^3}{6} + o(x^4)\
cos(x) = 1 - frac{x^2}{2} + o(x^4)
$$
https://en.wikipedia.org/wiki/Taylor_series#Trigonometric_functions
Just substitute functions with their expansions. Then just find lim of expression~polynomial/polynomial. Add more $x^n$ terms, if first 2 is not enough.
L'Hôpital's rule is another form of this more general approach.
answered Jan 8 at 18:30
Mike_Mike_
616
$endgroup$
One can evaluate all such lims using series expansions.
$$
sin(x) = x - frac{x^3}{6} + o(x^4)\
cos(x) = 1 - frac{x^2}{2} + o(x^4)
$$
https://en.wikipedia.org/wiki/Taylor_series#Trigonometric_functions
Just substitute functions with their expansions. Then just find lim of expression~polynomial/polynomial. Add more $x^n$ terms, if first 2 is not enough.
L'Hôpital's rule is another form of this more general approach.
answered Jan 8 at 18:30
Mike_Mike_
616
answered Jan 8 at 18:30
Mike_Mike_
616
answered Jan 8 at 18:30
Mike_Mike_
616
answered Jan 8 at 18:30
Mike_Mike_
616
616
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066530%2fhow-to-evaluate-lim-x-to-0-frac-sin2x-2-sinx43-cos2x-4-cosx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
