Using the Squeeze Theorem on $lim_{xto 0}frac{sin^2x}{x^2}$
$begingroup$
$$lim_{xto 0}frac{sin^2x}{x^2}$$
I'm trying to evaluate this limit using Squeeze Theorem. However, looking at the graph I know it approaches $1$, but I am getting $0$ using the Squeeze Theorem.
$$-frac{1}{x^2} < frac{sin^2x}{x^2} < frac{1}{x^2}$$
when I sub in $0$ it's just $0$. What am I doing wrong?
Edit: Wait, it's not zero! The upper and lower bounds are indeterminate. So I can't use squeeze theorem, correct?
limits
edited Dec 16 '18 at 18:58
Blue
49.1k870156
asked Dec 16 '18 at 18:52
J.W.J.W.
544
$endgroup$
add a comment |
$begingroup$
$$lim_{xto 0}frac{sin^2x}{x^2}$$
I'm trying to evaluate this limit using Squeeze Theorem. However, looking at the graph I know it approaches $1$, but I am getting $0$ using the Squeeze Theorem.
$$-frac{1}{x^2} < frac{sin^2x}{x^2} < frac{1}{x^2}$$
when I sub in $0$ it's just $0$. What am I doing wrong?
Edit: Wait, it's not zero! The upper and lower bounds are indeterminate. So I can't use squeeze theorem, correct?
limits
edited Dec 16 '18 at 18:58
Blue
49.1k870156
asked Dec 16 '18 at 18:52
J.W.J.W.
544
$endgroup$
1
$begingroup$
$1/x^2to +infty$
$endgroup$
– A.Γ.
Dec 16 '18 at 18:54
$begingroup$
You can only use squeeze theorem when upper and lower bound limits exist and are equal.
$endgroup$
– D.B.
Dec 16 '18 at 18:57
$begingroup$
Use L'Hopital's rule.
$endgroup$
– D.B.
Dec 16 '18 at 18:59
add a comment |
$begingroup$
$$lim_{xto 0}frac{sin^2x}{x^2}$$
I'm trying to evaluate this limit using Squeeze Theorem. However, looking at the graph I know it approaches $1$, but I am getting $0$ using the Squeeze Theorem.
$$-frac{1}{x^2} < frac{sin^2x}{x^2} < frac{1}{x^2}$$
when I sub in $0$ it's just $0$. What am I doing wrong?
Edit: Wait, it's not zero! The upper and lower bounds are indeterminate. So I can't use squeeze theorem, correct?
limits
edited Dec 16 '18 at 18:58
Blue
49.1k870156
asked Dec 16 '18 at 18:52
J.W.J.W.
544
$endgroup$
$$lim_{xto 0}frac{sin^2x}{x^2}$$
I'm trying to evaluate this limit using Squeeze Theorem. However, looking at the graph I know it approaches $1$, but I am getting $0$ using the Squeeze Theorem.
$$-frac{1}{x^2} < frac{sin^2x}{x^2} < frac{1}{x^2}$$
when I sub in $0$ it's just $0$. What am I doing wrong?
Edit: Wait, it's not zero! The upper and lower bounds are indeterminate. So I can't use squeeze theorem, correct?
limits
limits
edited Dec 16 '18 at 18:58
Blue
49.1k870156
asked Dec 16 '18 at 18:52
J.W.J.W.
544
edited Dec 16 '18 at 18:58
Blue
49.1k870156
asked Dec 16 '18 at 18:52
J.W.J.W.
544
edited Dec 16 '18 at 18:58
Blue
49.1k870156
edited Dec 16 '18 at 18:58
Blue
49.1k870156
edited Dec 16 '18 at 18:58
Blue
49.1k870156
49.1k870156
asked Dec 16 '18 at 18:52
J.W.J.W.
544
asked Dec 16 '18 at 18:52
J.W.J.W.
544
asked Dec 16 '18 at 18:52
J.W.J.W.
544
544
1
$begingroup$
$1/x^2to +infty$
$endgroup$
– A.Γ.
Dec 16 '18 at 18:54
$begingroup$
You can only use squeeze theorem when upper and lower bound limits exist and are equal.
$endgroup$
– D.B.
Dec 16 '18 at 18:57
$begingroup$
Use L'Hopital's rule.
$endgroup$
– D.B.
Dec 16 '18 at 18:59
add a comment |
1
$begingroup$
$1/x^2to +infty$
$endgroup$
– A.Γ.
Dec 16 '18 at 18:54
$begingroup$
You can only use squeeze theorem when upper and lower bound limits exist and are equal.
$endgroup$
– D.B.
Dec 16 '18 at 18:57
$begingroup$
Use L'Hopital's rule.
$endgroup$
– D.B.
Dec 16 '18 at 18:59
1
1
$begingroup$
$1/x^2to +infty$
$endgroup$
– A.Γ.
Dec 16 '18 at 18:54
$begingroup$
$1/x^2to +infty$
$endgroup$
– A.Γ.
Dec 16 '18 at 18:54
$begingroup$
You can only use squeeze theorem when upper and lower bound limits exist and are equal.
$endgroup$
– D.B.
Dec 16 '18 at 18:57
$begingroup$
You can only use squeeze theorem when upper and lower bound limits exist and are equal.
$endgroup$
– D.B.
Dec 16 '18 at 18:57
$begingroup$
Use L'Hopital's rule.
$endgroup$
– D.B.
Dec 16 '18 at 18:59
$begingroup$
Use L'Hopital's rule.
$endgroup$
– D.B.
Dec 16 '18 at 18:59
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The lower and upper bounds you write are right, but unfortunately the lower bound has limit $-infty$ and the upper bound has limit $infty$, so they can't be used to determine the given limit.
If you want to apply squeezing, you can prove geometrically that
$$
cos^2x<frac{sin^2x}{x^2}<frac{1}{cos^2x}
$$
which is basically the usual proof that
$$
lim_{xto0}frac{sin x}{x}=1
$$
answered Dec 16 '18 at 18:59
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
The function is even, we assume $0<x<1$.
By MVT,
$$sin(x)=xcos(c)$$ with $0<c<x$.
thus
$$cos^2(x)<frac{sin^2(x)}{x^2}=cos^2(c)<1$$
answered Dec 16 '18 at 19:02
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
$begingroup$
As you found out, the upper and lower bounds are indeterminate. Therefore, you cannot determine the limit that way. Instead, you can refer to the following well-known limit
$$lim_{x to 0}frac{sin x}{x} = 1$$
If you want to use the Squeeze Theorem, you can refer to the geometric proof of the limit above. (Then, you can apply the geometric proof here and reach a conclusion by the Squeeze Theorem.)
answered Dec 16 '18 at 19:03
KM101KM101
6,0901525
$endgroup$
add a comment |
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
});
}
});
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%2f3043000%2fusing-the-squeeze-theorem-on-lim-x-to-0-frac-sin2xx2%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
$begingroup$
The lower and upper bounds you write are right, but unfortunately the lower bound has limit $-infty$ and the upper bound has limit $infty$, so they can't be used to determine the given limit.
If you want to apply squeezing, you can prove geometrically that
$$
cos^2x<frac{sin^2x}{x^2}<frac{1}{cos^2x}
$$
which is basically the usual proof that
$$
lim_{xto0}frac{sin x}{x}=1
$$
answered Dec 16 '18 at 18:59
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
The lower and upper bounds you write are right, but unfortunately the lower bound has limit $-infty$ and the upper bound has limit $infty$, so they can't be used to determine the given limit.
If you want to apply squeezing, you can prove geometrically that
$$
cos^2x<frac{sin^2x}{x^2}<frac{1}{cos^2x}
$$
which is basically the usual proof that
$$
lim_{xto0}frac{sin x}{x}=1
$$
answered Dec 16 '18 at 18:59
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
The lower and upper bounds you write are right, but unfortunately the lower bound has limit $-infty$ and the upper bound has limit $infty$, so they can't be used to determine the given limit.
If you want to apply squeezing, you can prove geometrically that
$$
cos^2x<frac{sin^2x}{x^2}<frac{1}{cos^2x}
$$
which is basically the usual proof that
$$
lim_{xto0}frac{sin x}{x}=1
$$
answered Dec 16 '18 at 18:59
egregegreg
184k1486205
$endgroup$
The lower and upper bounds you write are right, but unfortunately the lower bound has limit $-infty$ and the upper bound has limit $infty$, so they can't be used to determine the given limit.
If you want to apply squeezing, you can prove geometrically that
$$
cos^2x<frac{sin^2x}{x^2}<frac{1}{cos^2x}
$$
which is basically the usual proof that
$$
lim_{xto0}frac{sin x}{x}=1
$$
answered Dec 16 '18 at 18:59
egregegreg
184k1486205
answered Dec 16 '18 at 18:59
egregegreg
184k1486205
answered Dec 16 '18 at 18:59
egregegreg
184k1486205
answered Dec 16 '18 at 18:59
egregegreg
184k1486205
184k1486205
add a comment |
add a comment |
$begingroup$
The function is even, we assume $0<x<1$.
By MVT,
$$sin(x)=xcos(c)$$ with $0<c<x$.
thus
$$cos^2(x)<frac{sin^2(x)}{x^2}=cos^2(c)<1$$
answered Dec 16 '18 at 19:02
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
$begingroup$
The function is even, we assume $0<x<1$.
By MVT,
$$sin(x)=xcos(c)$$ with $0<c<x$.
thus
$$cos^2(x)<frac{sin^2(x)}{x^2}=cos^2(c)<1$$
answered Dec 16 '18 at 19:02
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
$begingroup$
The function is even, we assume $0<x<1$.
By MVT,
$$sin(x)=xcos(c)$$ with $0<c<x$.
thus
$$cos^2(x)<frac{sin^2(x)}{x^2}=cos^2(c)<1$$
answered Dec 16 '18 at 19:02
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
The function is even, we assume $0<x<1$.
By MVT,
$$sin(x)=xcos(c)$$ with $0<c<x$.
thus
$$cos^2(x)<frac{sin^2(x)}{x^2}=cos^2(c)<1$$
answered Dec 16 '18 at 19:02
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Dec 16 '18 at 19:02
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Dec 16 '18 at 19:02
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Dec 16 '18 at 19:02
hamam_Abdallahhamam_Abdallah
38.2k21634
38.2k21634
add a comment |
add a comment |
$begingroup$
As you found out, the upper and lower bounds are indeterminate. Therefore, you cannot determine the limit that way. Instead, you can refer to the following well-known limit
$$lim_{x to 0}frac{sin x}{x} = 1$$
If you want to use the Squeeze Theorem, you can refer to the geometric proof of the limit above. (Then, you can apply the geometric proof here and reach a conclusion by the Squeeze Theorem.)
answered Dec 16 '18 at 19:03
KM101KM101
6,0901525
$endgroup$
add a comment |
$begingroup$
As you found out, the upper and lower bounds are indeterminate. Therefore, you cannot determine the limit that way. Instead, you can refer to the following well-known limit
$$lim_{x to 0}frac{sin x}{x} = 1$$
If you want to use the Squeeze Theorem, you can refer to the geometric proof of the limit above. (Then, you can apply the geometric proof here and reach a conclusion by the Squeeze Theorem.)
answered Dec 16 '18 at 19:03
KM101KM101
6,0901525
$endgroup$
add a comment |
$begingroup$
As you found out, the upper and lower bounds are indeterminate. Therefore, you cannot determine the limit that way. Instead, you can refer to the following well-known limit
$$lim_{x to 0}frac{sin x}{x} = 1$$
If you want to use the Squeeze Theorem, you can refer to the geometric proof of the limit above. (Then, you can apply the geometric proof here and reach a conclusion by the Squeeze Theorem.)
answered Dec 16 '18 at 19:03
KM101KM101
6,0901525
$endgroup$
As you found out, the upper and lower bounds are indeterminate. Therefore, you cannot determine the limit that way. Instead, you can refer to the following well-known limit
$$lim_{x to 0}frac{sin x}{x} = 1$$
If you want to use the Squeeze Theorem, you can refer to the geometric proof of the limit above. (Then, you can apply the geometric proof here and reach a conclusion by the Squeeze Theorem.)
answered Dec 16 '18 at 19:03
KM101KM101
6,0901525
edited Dec 16 '18 at 19:15
answered Dec 16 '18 at 19:03
KM101KM101
6,0901525
answered Dec 16 '18 at 19:03
KM101KM101
6,0901525
answered Dec 16 '18 at 19:03
KM101KM101
6,0901525
6,0901525
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%2f3043000%2fusing-the-squeeze-theorem-on-lim-x-to-0-frac-sin2xx2%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
1
$begingroup$
$1/x^2to +infty$
$endgroup$
– A.Γ.
Dec 16 '18 at 18:54
$begingroup$
You can only use squeeze theorem when upper and lower bound limits exist and are equal.
$endgroup$
– D.B.
Dec 16 '18 at 18:57
$begingroup$
Use L'Hopital's rule.
$endgroup$
– D.B.
Dec 16 '18 at 18:59