Find the limit of $frac{1}{1-cos(x)}-frac{2}{x^2}$ as $x$ approaches $0$
up vote
0
down vote
favorite
I need to find $$lim_{xto 0}left(frac{1}{1-cos(x)}-frac{2}{x^2}right)$$
I already found it using Taylor series. However, I'm looking for a solution without Taylor series expansion or L'Hopital's rule because the problem was given in a calculus class at a point when only limits had been studied.
calculus limits limits-without-lhopital
asked Nov 17 at 21:47
bjorn93
325
|
show 5 more comments
up vote
0
down vote
favorite
I need to find $$lim_{xto 0}left(frac{1}{1-cos(x)}-frac{2}{x^2}right)$$
I already found it using Taylor series. However, I'm looking for a solution without Taylor series expansion or L'Hopital's rule because the problem was given in a calculus class at a point when only limits had been studied.
calculus limits limits-without-lhopital
asked Nov 17 at 21:47
bjorn93
325
It is a bit strange that this problem has been given at this stage since it seems not solveble without Taylor, lHopital or derivatives concept.
– gimusi
Nov 17 at 22:01
1
See the related OP
– gimusi
Nov 17 at 22:03
@KeyFlex It is not a duplicate because the OP is looking for a solution without Taylor.
– gimusi
Nov 17 at 22:17
The limit involves evaluating $lim_{xto0}frac{x-sin x}{x^3}$ which could be done without Taylor or l'Hôpital, but in very convoluted ways.
– egreg
Nov 17 at 22:19
@egreg Are you sure it can be done using $lim_{xto0}frac{x-sin x}{x^3}=frac16$?
– gimusi
Nov 17 at 22:22
|
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I need to find $$lim_{xto 0}left(frac{1}{1-cos(x)}-frac{2}{x^2}right)$$
I already found it using Taylor series. However, I'm looking for a solution without Taylor series expansion or L'Hopital's rule because the problem was given in a calculus class at a point when only limits had been studied.
calculus limits limits-without-lhopital
asked Nov 17 at 21:47
bjorn93
325
I need to find $$lim_{xto 0}left(frac{1}{1-cos(x)}-frac{2}{x^2}right)$$
I already found it using Taylor series. However, I'm looking for a solution without Taylor series expansion or L'Hopital's rule because the problem was given in a calculus class at a point when only limits had been studied.
calculus limits limits-without-lhopital
calculus limits limits-without-lhopital
asked Nov 17 at 21:47
bjorn93
325
asked Nov 17 at 21:47
bjorn93
325
asked Nov 17 at 21:47
bjorn93
325
asked Nov 17 at 21:47
bjorn93
325
asked Nov 17 at 21:47
bjorn93
325
325
It is a bit strange that this problem has been given at this stage since it seems not solveble without Taylor, lHopital or derivatives concept.
– gimusi
Nov 17 at 22:01
1
See the related OP
– gimusi
Nov 17 at 22:03
@KeyFlex It is not a duplicate because the OP is looking for a solution without Taylor.
– gimusi
Nov 17 at 22:17
The limit involves evaluating $lim_{xto0}frac{x-sin x}{x^3}$ which could be done without Taylor or l'Hôpital, but in very convoluted ways.
– egreg
Nov 17 at 22:19
@egreg Are you sure it can be done using $lim_{xto0}frac{x-sin x}{x^3}=frac16$?
– gimusi
Nov 17 at 22:22
|
show 5 more comments
It is a bit strange that this problem has been given at this stage since it seems not solveble without Taylor, lHopital or derivatives concept.
– gimusi
Nov 17 at 22:01
1
See the related OP
– gimusi
Nov 17 at 22:03
@KeyFlex It is not a duplicate because the OP is looking for a solution without Taylor.
– gimusi
Nov 17 at 22:17
The limit involves evaluating $lim_{xto0}frac{x-sin x}{x^3}$ which could be done without Taylor or l'Hôpital, but in very convoluted ways.
– egreg
Nov 17 at 22:19
@egreg Are you sure it can be done using $lim_{xto0}frac{x-sin x}{x^3}=frac16$?
– gimusi
Nov 17 at 22:22
It is a bit strange that this problem has been given at this stage since it seems not solveble without Taylor, lHopital or derivatives concept.
– gimusi
Nov 17 at 22:01
It is a bit strange that this problem has been given at this stage since it seems not solveble without Taylor, lHopital or derivatives concept.
– gimusi
Nov 17 at 22:01
1
1
See the related OP
– gimusi
Nov 17 at 22:03
See the related OP
– gimusi
Nov 17 at 22:03
@KeyFlex It is not a duplicate because the OP is looking for a solution without Taylor.
– gimusi
Nov 17 at 22:17
@KeyFlex It is not a duplicate because the OP is looking for a solution without Taylor.
– gimusi
Nov 17 at 22:17
The limit involves evaluating $lim_{xto0}frac{x-sin x}{x^3}$ which could be done without Taylor or l'Hôpital, but in very convoluted ways.
– egreg
Nov 17 at 22:19
The limit involves evaluating $lim_{xto0}frac{x-sin x}{x^3}$ which could be done without Taylor or l'Hôpital, but in very convoluted ways.
– egreg
Nov 17 at 22:19
@egreg Are you sure it can be done using $lim_{xto0}frac{x-sin x}{x^3}=frac16$?
– gimusi
Nov 17 at 22:22
@egreg Are you sure it can be done using $lim_{xto0}frac{x-sin x}{x^3}=frac16$?
– gimusi
Nov 17 at 22:22
|
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
With the substitution $x=2z$, the limit becomes
$$
lim_{zto0}left(frac{1}{sin^2z}-frac{1}{z^2}right)=
frac{1}{2}lim_{zto0}frac{z^2-sin^2z}{z^2sin^2z}=
frac{1}{2}lim_{zto0}frac{z-sin z}{z^3}frac{z+sin z}{z}frac{z^2}{sin^2z}
$$
(see https://math.stackexchange.com/a/1357590/62967 for the idea about the substitution, not for the complete solution, that uses Taylor).
The second and third fractions have elementary limits $2$ and $1$ respectively. For the first fraction refer to https://math.stackexchange.com/a/1337564/62967
answered Nov 17 at 22:44
egreg
175k1383198
The limits that are being used make use of Taylor expansions in their derivations.
– herb steinberg
Nov 17 at 22:47
@herbsteinberg No it suffices to use $$frac{z^2-sin^2z}{z^2sin^2z}=frac{z^2-sin^2z}{z^4}frac{z^2}{sin^2z}$$ and use that $frac{z^2}{sin^2z}to 1$.
– gimusi
Nov 17 at 22:50
@herbsteinberg Did you read my answer? I'm following Jack up to the point where he uses Taylor, but I'm pointing to another strategy for computing the limit of the first fraction without Taylor. The second fraction is $1+frac{sin z}{z}$, which doesn't need Taylor.
– egreg
Nov 17 at 22:50
The first fraction was my point of contention. I looked up the reference, and I have to agree with you. I presume $frac{sinz}{z}to 1$ can be derived without Taylor.
– herb steinberg
Nov 17 at 23:52
@herbsteinberg When “no l'Hôpital” is requested, some limits have to be allowed and $sin z/z$ is one of them.
– egreg
Nov 18 at 9:49
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
With the substitution $x=2z$, the limit becomes
$$
lim_{zto0}left(frac{1}{sin^2z}-frac{1}{z^2}right)=
frac{1}{2}lim_{zto0}frac{z^2-sin^2z}{z^2sin^2z}=
frac{1}{2}lim_{zto0}frac{z-sin z}{z^3}frac{z+sin z}{z}frac{z^2}{sin^2z}
$$
(see https://math.stackexchange.com/a/1357590/62967 for the idea about the substitution, not for the complete solution, that uses Taylor).
The second and third fractions have elementary limits $2$ and $1$ respectively. For the first fraction refer to https://math.stackexchange.com/a/1337564/62967
answered Nov 17 at 22:44
egreg
175k1383198
The limits that are being used make use of Taylor expansions in their derivations.
– herb steinberg
Nov 17 at 22:47
@herbsteinberg No it suffices to use $$frac{z^2-sin^2z}{z^2sin^2z}=frac{z^2-sin^2z}{z^4}frac{z^2}{sin^2z}$$ and use that $frac{z^2}{sin^2z}to 1$.
– gimusi
Nov 17 at 22:50
@herbsteinberg Did you read my answer? I'm following Jack up to the point where he uses Taylor, but I'm pointing to another strategy for computing the limit of the first fraction without Taylor. The second fraction is $1+frac{sin z}{z}$, which doesn't need Taylor.
– egreg
Nov 17 at 22:50
The first fraction was my point of contention. I looked up the reference, and I have to agree with you. I presume $frac{sinz}{z}to 1$ can be derived without Taylor.
– herb steinberg
Nov 17 at 23:52
@herbsteinberg When “no l'Hôpital” is requested, some limits have to be allowed and $sin z/z$ is one of them.
– egreg
Nov 18 at 9:49
|
show 2 more comments
up vote
5
down vote
accepted
With the substitution $x=2z$, the limit becomes
$$
lim_{zto0}left(frac{1}{sin^2z}-frac{1}{z^2}right)=
frac{1}{2}lim_{zto0}frac{z^2-sin^2z}{z^2sin^2z}=
frac{1}{2}lim_{zto0}frac{z-sin z}{z^3}frac{z+sin z}{z}frac{z^2}{sin^2z}
$$
(see https://math.stackexchange.com/a/1357590/62967 for the idea about the substitution, not for the complete solution, that uses Taylor).
The second and third fractions have elementary limits $2$ and $1$ respectively. For the first fraction refer to https://math.stackexchange.com/a/1337564/62967
answered Nov 17 at 22:44
egreg
175k1383198
The limits that are being used make use of Taylor expansions in their derivations.
– herb steinberg
Nov 17 at 22:47
@herbsteinberg No it suffices to use $$frac{z^2-sin^2z}{z^2sin^2z}=frac{z^2-sin^2z}{z^4}frac{z^2}{sin^2z}$$ and use that $frac{z^2}{sin^2z}to 1$.
– gimusi
Nov 17 at 22:50
@herbsteinberg Did you read my answer? I'm following Jack up to the point where he uses Taylor, but I'm pointing to another strategy for computing the limit of the first fraction without Taylor. The second fraction is $1+frac{sin z}{z}$, which doesn't need Taylor.
– egreg
Nov 17 at 22:50
The first fraction was my point of contention. I looked up the reference, and I have to agree with you. I presume $frac{sinz}{z}to 1$ can be derived without Taylor.
– herb steinberg
Nov 17 at 23:52
@herbsteinberg When “no l'Hôpital” is requested, some limits have to be allowed and $sin z/z$ is one of them.
– egreg
Nov 18 at 9:49
|
show 2 more comments
up vote
5
down vote
accepted
up vote
5
down vote
accepted
With the substitution $x=2z$, the limit becomes
$$
lim_{zto0}left(frac{1}{sin^2z}-frac{1}{z^2}right)=
frac{1}{2}lim_{zto0}frac{z^2-sin^2z}{z^2sin^2z}=
frac{1}{2}lim_{zto0}frac{z-sin z}{z^3}frac{z+sin z}{z}frac{z^2}{sin^2z}
$$
(see https://math.stackexchange.com/a/1357590/62967 for the idea about the substitution, not for the complete solution, that uses Taylor).
The second and third fractions have elementary limits $2$ and $1$ respectively. For the first fraction refer to https://math.stackexchange.com/a/1337564/62967
answered Nov 17 at 22:44
egreg
175k1383198
With the substitution $x=2z$, the limit becomes
$$
lim_{zto0}left(frac{1}{sin^2z}-frac{1}{z^2}right)=
frac{1}{2}lim_{zto0}frac{z^2-sin^2z}{z^2sin^2z}=
frac{1}{2}lim_{zto0}frac{z-sin z}{z^3}frac{z+sin z}{z}frac{z^2}{sin^2z}
$$
(see https://math.stackexchange.com/a/1357590/62967 for the idea about the substitution, not for the complete solution, that uses Taylor).
The second and third fractions have elementary limits $2$ and $1$ respectively. For the first fraction refer to https://math.stackexchange.com/a/1337564/62967
answered Nov 17 at 22:44
egreg
175k1383198
edited Nov 17 at 22:53
answered Nov 17 at 22:44
egreg
175k1383198
answered Nov 17 at 22:44
egreg
175k1383198
answered Nov 17 at 22:44
egreg
175k1383198
175k1383198
The limits that are being used make use of Taylor expansions in their derivations.
– herb steinberg
Nov 17 at 22:47
@herbsteinberg No it suffices to use $$frac{z^2-sin^2z}{z^2sin^2z}=frac{z^2-sin^2z}{z^4}frac{z^2}{sin^2z}$$ and use that $frac{z^2}{sin^2z}to 1$.
– gimusi
Nov 17 at 22:50
@herbsteinberg Did you read my answer? I'm following Jack up to the point where he uses Taylor, but I'm pointing to another strategy for computing the limit of the first fraction without Taylor. The second fraction is $1+frac{sin z}{z}$, which doesn't need Taylor.
– egreg
Nov 17 at 22:50
The first fraction was my point of contention. I looked up the reference, and I have to agree with you. I presume $frac{sinz}{z}to 1$ can be derived without Taylor.
– herb steinberg
Nov 17 at 23:52
@herbsteinberg When “no l'Hôpital” is requested, some limits have to be allowed and $sin z/z$ is one of them.
– egreg
Nov 18 at 9:49
|
show 2 more comments
The limits that are being used make use of Taylor expansions in their derivations.
– herb steinberg
Nov 17 at 22:47
@herbsteinberg No it suffices to use $$frac{z^2-sin^2z}{z^2sin^2z}=frac{z^2-sin^2z}{z^4}frac{z^2}{sin^2z}$$ and use that $frac{z^2}{sin^2z}to 1$.
– gimusi
Nov 17 at 22:50
@herbsteinberg Did you read my answer? I'm following Jack up to the point where he uses Taylor, but I'm pointing to another strategy for computing the limit of the first fraction without Taylor. The second fraction is $1+frac{sin z}{z}$, which doesn't need Taylor.
– egreg
Nov 17 at 22:50
The first fraction was my point of contention. I looked up the reference, and I have to agree with you. I presume $frac{sinz}{z}to 1$ can be derived without Taylor.
– herb steinberg
Nov 17 at 23:52
@herbsteinberg When “no l'Hôpital” is requested, some limits have to be allowed and $sin z/z$ is one of them.
– egreg
Nov 18 at 9:49
The limits that are being used make use of Taylor expansions in their derivations.
– herb steinberg
Nov 17 at 22:47
The limits that are being used make use of Taylor expansions in their derivations.
– herb steinberg
Nov 17 at 22:47
@herbsteinberg No it suffices to use $$frac{z^2-sin^2z}{z^2sin^2z}=frac{z^2-sin^2z}{z^4}frac{z^2}{sin^2z}$$ and use that $frac{z^2}{sin^2z}to 1$.
– gimusi
Nov 17 at 22:50
@herbsteinberg No it suffices to use $$frac{z^2-sin^2z}{z^2sin^2z}=frac{z^2-sin^2z}{z^4}frac{z^2}{sin^2z}$$ and use that $frac{z^2}{sin^2z}to 1$.
– gimusi
Nov 17 at 22:50
@herbsteinberg Did you read my answer? I'm following Jack up to the point where he uses Taylor, but I'm pointing to another strategy for computing the limit of the first fraction without Taylor. The second fraction is $1+frac{sin z}{z}$, which doesn't need Taylor.
– egreg
Nov 17 at 22:50
@herbsteinberg Did you read my answer? I'm following Jack up to the point where he uses Taylor, but I'm pointing to another strategy for computing the limit of the first fraction without Taylor. The second fraction is $1+frac{sin z}{z}$, which doesn't need Taylor.
– egreg
Nov 17 at 22:50
The first fraction was my point of contention. I looked up the reference, and I have to agree with you. I presume $frac{sinz}{z}to 1$ can be derived without Taylor.
– herb steinberg
Nov 17 at 23:52
The first fraction was my point of contention. I looked up the reference, and I have to agree with you. I presume $frac{sinz}{z}to 1$ can be derived without Taylor.
– herb steinberg
Nov 17 at 23:52
@herbsteinberg When “no l'Hôpital” is requested, some limits have to be allowed and $sin z/z$ is one of them.
– egreg
Nov 18 at 9:49
@herbsteinberg When “no l'Hôpital” is requested, some limits have to be allowed and $sin z/z$ is one of them.
– egreg
Nov 18 at 9:49
|
show 2 more comments
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.
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%2f3002853%2ffind-the-limit-of-frac11-cosx-frac2x2-as-x-approaches-0%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
It is a bit strange that this problem has been given at this stage since it seems not solveble without Taylor, lHopital or derivatives concept.
– gimusi
Nov 17 at 22:01
1
See the related OP
– gimusi
Nov 17 at 22:03
@KeyFlex It is not a duplicate because the OP is looking for a solution without Taylor.
– gimusi
Nov 17 at 22:17
The limit involves evaluating $lim_{xto0}frac{x-sin x}{x^3}$ which could be done without Taylor or l'Hôpital, but in very convoluted ways.
– egreg
Nov 17 at 22:19
@egreg Are you sure it can be done using $lim_{xto0}frac{x-sin x}{x^3}=frac16$?
– gimusi
Nov 17 at 22:22