A solution for a limit
$begingroup$
What do you think about my solution on this limit:
$lim_{xto 0}frac{2x-sin(2x)}{x^3}=frac{4}{3}$
We make the substitution $x=3y$
$$lim_{xrightarrow 0}frac{2x-sin 2x}{x^{3}}=lim_{yrightarrow 0}frac{6y-sin 6y}{27y^{3}}=lim_{yrightarrow 0}frac{6y-3sin 2y}{27y^{3}}+lim_{yrightarrow 0}frac{4(sin 2y)^{3}}{27{y^{3}}}$$
We get
$$l=frac{3}{27}l+frac{32}{27}$$
and
$$l=frac{4}{3}$$
limits
asked Jan 9 at 14:40
G. VasiG. Vasi
611
$endgroup$
add a comment |
$begingroup$
What do you think about my solution on this limit:
$lim_{xto 0}frac{2x-sin(2x)}{x^3}=frac{4}{3}$
We make the substitution $x=3y$
$$lim_{xrightarrow 0}frac{2x-sin 2x}{x^{3}}=lim_{yrightarrow 0}frac{6y-sin 6y}{27y^{3}}=lim_{yrightarrow 0}frac{6y-3sin 2y}{27y^{3}}+lim_{yrightarrow 0}frac{4(sin 2y)^{3}}{27{y^{3}}}$$
We get
$$l=frac{3}{27}l+frac{32}{27}$$
and
$$l=frac{4}{3}$$
limits
asked Jan 9 at 14:40
G. VasiG. Vasi
611
$endgroup$
1
$begingroup$
You did not prove that there was a limit. It is fine apart from that.
$endgroup$
– Mindlack
Jan 9 at 14:43
$begingroup$
How can you write $sin 6y=3sin 2y$?
$endgroup$
– pie314271
Jan 9 at 14:44
$begingroup$
@pie314271 He's using the triple angle formula for $sin$
$endgroup$
– saulspatz
Jan 9 at 14:45
$begingroup$
(Oh, I thought that the plus sign at the end of the first line was another equals sign for some reason. Sorry.)
$endgroup$
– pie314271
Jan 9 at 14:47
$begingroup$
math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:06
add a comment |
$begingroup$
What do you think about my solution on this limit:
$lim_{xto 0}frac{2x-sin(2x)}{x^3}=frac{4}{3}$
We make the substitution $x=3y$
$$lim_{xrightarrow 0}frac{2x-sin 2x}{x^{3}}=lim_{yrightarrow 0}frac{6y-sin 6y}{27y^{3}}=lim_{yrightarrow 0}frac{6y-3sin 2y}{27y^{3}}+lim_{yrightarrow 0}frac{4(sin 2y)^{3}}{27{y^{3}}}$$
We get
$$l=frac{3}{27}l+frac{32}{27}$$
and
$$l=frac{4}{3}$$
limits
asked Jan 9 at 14:40
G. VasiG. Vasi
611
$endgroup$
What do you think about my solution on this limit:
$lim_{xto 0}frac{2x-sin(2x)}{x^3}=frac{4}{3}$
We make the substitution $x=3y$
$$lim_{xrightarrow 0}frac{2x-sin 2x}{x^{3}}=lim_{yrightarrow 0}frac{6y-sin 6y}{27y^{3}}=lim_{yrightarrow 0}frac{6y-3sin 2y}{27y^{3}}+lim_{yrightarrow 0}frac{4(sin 2y)^{3}}{27{y^{3}}}$$
We get
$$l=frac{3}{27}l+frac{32}{27}$$
and
$$l=frac{4}{3}$$
limits
limits
asked Jan 9 at 14:40
G. VasiG. Vasi
611
asked Jan 9 at 14:40
G. VasiG. Vasi
611
asked Jan 9 at 14:40
G. VasiG. Vasi
611
asked Jan 9 at 14:40
G. VasiG. Vasi
611
asked Jan 9 at 14:40
G. VasiG. Vasi
611
611
1
$begingroup$
You did not prove that there was a limit. It is fine apart from that.
$endgroup$
– Mindlack
Jan 9 at 14:43
$begingroup$
How can you write $sin 6y=3sin 2y$?
$endgroup$
– pie314271
Jan 9 at 14:44
$begingroup$
@pie314271 He's using the triple angle formula for $sin$
$endgroup$
– saulspatz
Jan 9 at 14:45
$begingroup$
(Oh, I thought that the plus sign at the end of the first line was another equals sign for some reason. Sorry.)
$endgroup$
– pie314271
Jan 9 at 14:47
$begingroup$
math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:06
add a comment |
1
$begingroup$
You did not prove that there was a limit. It is fine apart from that.
$endgroup$
– Mindlack
Jan 9 at 14:43
$begingroup$
How can you write $sin 6y=3sin 2y$?
$endgroup$
– pie314271
Jan 9 at 14:44
$begingroup$
@pie314271 He's using the triple angle formula for $sin$
$endgroup$
– saulspatz
Jan 9 at 14:45
$begingroup$
(Oh, I thought that the plus sign at the end of the first line was another equals sign for some reason. Sorry.)
$endgroup$
– pie314271
Jan 9 at 14:47
$begingroup$
math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:06
1
1
$begingroup$
You did not prove that there was a limit. It is fine apart from that.
$endgroup$
– Mindlack
Jan 9 at 14:43
$begingroup$
You did not prove that there was a limit. It is fine apart from that.
$endgroup$
– Mindlack
Jan 9 at 14:43
$begingroup$
How can you write $sin 6y=3sin 2y$?
$endgroup$
– pie314271
Jan 9 at 14:44
$begingroup$
How can you write $sin 6y=3sin 2y$?
$endgroup$
– pie314271
Jan 9 at 14:44
$begingroup$
@pie314271 He's using the triple angle formula for $sin$
$endgroup$
– saulspatz
Jan 9 at 14:45
$begingroup$
@pie314271 He's using the triple angle formula for $sin$
$endgroup$
– saulspatz
Jan 9 at 14:45
$begingroup$
(Oh, I thought that the plus sign at the end of the first line was another equals sign for some reason. Sorry.)
$endgroup$
– pie314271
Jan 9 at 14:47
$begingroup$
(Oh, I thought that the plus sign at the end of the first line was another equals sign for some reason. Sorry.)
$endgroup$
– pie314271
Jan 9 at 14:47
$begingroup$
math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:06
$begingroup$
math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This looks fine, except that you didn't prove your original limit exists. But this is a rather complicated way to derive this limit, you could have simply used L'Hospital's rule twice, or a simple Taylor expansion in the form $sin(2x) = 2x - frac{(2x)^3}{6} + o(x^3)$.
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1587
$endgroup$
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%2f3067515%2fa-solution-for-a-limit%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This looks fine, except that you didn't prove your original limit exists. But this is a rather complicated way to derive this limit, you could have simply used L'Hospital's rule twice, or a simple Taylor expansion in the form $sin(2x) = 2x - frac{(2x)^3}{6} + o(x^3)$.
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1587
$endgroup$
add a comment |
$begingroup$
This looks fine, except that you didn't prove your original limit exists. But this is a rather complicated way to derive this limit, you could have simply used L'Hospital's rule twice, or a simple Taylor expansion in the form $sin(2x) = 2x - frac{(2x)^3}{6} + o(x^3)$.
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1587
$endgroup$
add a comment |
$begingroup$
This looks fine, except that you didn't prove your original limit exists. But this is a rather complicated way to derive this limit, you could have simply used L'Hospital's rule twice, or a simple Taylor expansion in the form $sin(2x) = 2x - frac{(2x)^3}{6} + o(x^3)$.
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1587
$endgroup$
This looks fine, except that you didn't prove your original limit exists. But this is a rather complicated way to derive this limit, you could have simply used L'Hospital's rule twice, or a simple Taylor expansion in the form $sin(2x) = 2x - frac{(2x)^3}{6} + o(x^3)$.
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1587
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1587
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1587
answered Jan 9 at 14:50
A. BailleulA. Bailleul
1587
1587
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%2f3067515%2fa-solution-for-a-limit%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
$begingroup$
You did not prove that there was a limit. It is fine apart from that.
$endgroup$
– Mindlack
Jan 9 at 14:43
$begingroup$
How can you write $sin 6y=3sin 2y$?
$endgroup$
– pie314271
Jan 9 at 14:44
$begingroup$
@pie314271 He's using the triple angle formula for $sin$
$endgroup$
– saulspatz
Jan 9 at 14:45
$begingroup$
(Oh, I thought that the plus sign at the end of the first line was another equals sign for some reason. Sorry.)
$endgroup$
– pie314271
Jan 9 at 14:47
$begingroup$
math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:06