Why is $lim_{mto infty} frac{sinleft(frac{pi}{2^m}right)}{frac{pi}{2^m}} = 1$?
I am currently working on a proof with a good friend of mine that involves adding more and more triangles to the sides of a regular polygon but keeping the longest diagonal constant until eventually, it becomes a circle. And we ended up with this formula.
$4$-sided regular→$8$-sided regular→$16$-sided regular→$32$-sided regular$to ldots to n$-sided regular
(When $n$ tends to infinity, the area will be equal to that of a circle with the longest diagonal as diameter)
The Question is:
Can someone explain in detail:
Why does $$frac {sin frac {pi}{2^m}}{frac{pi}{2^m}} = 1$$ when $m$ tends to infinity?
Thanks in advance :D
Note I am still a beginner in finding limits. It would help greatly if you can explain step by step.
limits trigonometry
edited yesterday
David K
52.7k340115
asked yesterday
dfd
1
New contributor
dfd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I am currently working on a proof with a good friend of mine that involves adding more and more triangles to the sides of a regular polygon but keeping the longest diagonal constant until eventually, it becomes a circle. And we ended up with this formula.
$4$-sided regular→$8$-sided regular→$16$-sided regular→$32$-sided regular$to ldots to n$-sided regular
(When $n$ tends to infinity, the area will be equal to that of a circle with the longest diagonal as diameter)
The Question is:
Can someone explain in detail:
Why does $$frac {sin frac {pi}{2^m}}{frac{pi}{2^m}} = 1$$ when $m$ tends to infinity?
Thanks in advance :D
Note I am still a beginner in finding limits. It would help greatly if you can explain step by step.
limits trigonometry
edited yesterday
David K
52.7k340115
asked yesterday
dfd
1
New contributor
dfd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I am currently working on a proof with a good friend of mine that involves adding more and more triangles to the sides of a regular polygon but keeping the longest diagonal constant until eventually, it becomes a circle. And we ended up with this formula.
$4$-sided regular→$8$-sided regular→$16$-sided regular→$32$-sided regular$to ldots to n$-sided regular
(When $n$ tends to infinity, the area will be equal to that of a circle with the longest diagonal as diameter)
The Question is:
Can someone explain in detail:
Why does $$frac {sin frac {pi}{2^m}}{frac{pi}{2^m}} = 1$$ when $m$ tends to infinity?
Thanks in advance :D
Note I am still a beginner in finding limits. It would help greatly if you can explain step by step.
limits trigonometry
edited yesterday
David K
52.7k340115
asked yesterday
dfd
1
New contributor
dfd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am currently working on a proof with a good friend of mine that involves adding more and more triangles to the sides of a regular polygon but keeping the longest diagonal constant until eventually, it becomes a circle. And we ended up with this formula.
$4$-sided regular→$8$-sided regular→$16$-sided regular→$32$-sided regular$to ldots to n$-sided regular
(When $n$ tends to infinity, the area will be equal to that of a circle with the longest diagonal as diameter)
The Question is:
Can someone explain in detail:
Why does $$frac {sin frac {pi}{2^m}}{frac{pi}{2^m}} = 1$$ when $m$ tends to infinity?
Thanks in advance :D
Note I am still a beginner in finding limits. It would help greatly if you can explain step by step.
limits trigonometry
limits trigonometry
edited yesterday
David K
52.7k340115
asked yesterday
dfd
1
New contributor
dfd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
David K
52.7k340115
asked yesterday
dfd
1
New contributor
dfd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
David K
52.7k340115
edited yesterday
David K
52.7k340115
edited yesterday
David K
52.7k340115
52.7k340115
asked yesterday
dfd
1
New contributor
dfd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
dfd
1
asked yesterday
dfd
1
1
New contributor
dfd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
dfd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
dfd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Because we have
$$lim_{x to 0}frac{sin x}{x} = 1$$
which is a standard limit typically proven through simple geometrical observations followed by the Squeeze Theorem. You can check this link for a various explanation of why this works, particularly the one by robjohn, which I find the simplest and most elegant.
Another way of stating the above limit is
$$sin x sim x; quad x to 0$$
(Obviously, if $frac{sin x}{x} to 1$ as $x to 0$, then $sin x$ becomes very close to $x$.)
In case of your limit, you have
$$lim_{m to infty} frac{sinleft(frac{pi}{2^m}right)}{frac{pi}{2^m}}$$
It’s clear that as $m to infty$, $frac{pi}{2^m} to 0$ due to the growing denominator, so we make a substitution $t = frac{pi}{2^m}$. Clearly, $t to 0$ as $m to infty$, so the limit becomes
$$lim_{t to 0}frac{sin t}{t}$$
which becomes $1$.
answered yesterday
KM101
5,3691423
add a comment |
If you replace $fracpi{2^m}$ with x, it is obvious that $x$ goes to zero when $m$ goes to infinity. So your limit is equivalent to:
$$lim_{x to 0}frac{sin x}{x}$$
...which is 1. If you learn to evaluate just one limit, this is the one :) You can easily find several elementary proofs on the web and I really recommend this one from MSE.
answered yesterday
Oldboy
7,1191832
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
});
}
});
dfd is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3060629%2fwhy-is-lim-m-to-infty-frac-sin-left-frac-pi2m-right-frac-pi2%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
Because we have
$$lim_{x to 0}frac{sin x}{x} = 1$$
which is a standard limit typically proven through simple geometrical observations followed by the Squeeze Theorem. You can check this link for a various explanation of why this works, particularly the one by robjohn, which I find the simplest and most elegant.
Another way of stating the above limit is
$$sin x sim x; quad x to 0$$
(Obviously, if $frac{sin x}{x} to 1$ as $x to 0$, then $sin x$ becomes very close to $x$.)
In case of your limit, you have
$$lim_{m to infty} frac{sinleft(frac{pi}{2^m}right)}{frac{pi}{2^m}}$$
It’s clear that as $m to infty$, $frac{pi}{2^m} to 0$ due to the growing denominator, so we make a substitution $t = frac{pi}{2^m}$. Clearly, $t to 0$ as $m to infty$, so the limit becomes
$$lim_{t to 0}frac{sin t}{t}$$
which becomes $1$.
answered yesterday
KM101
5,3691423
add a comment |
Because we have
$$lim_{x to 0}frac{sin x}{x} = 1$$
which is a standard limit typically proven through simple geometrical observations followed by the Squeeze Theorem. You can check this link for a various explanation of why this works, particularly the one by robjohn, which I find the simplest and most elegant.
Another way of stating the above limit is
$$sin x sim x; quad x to 0$$
(Obviously, if $frac{sin x}{x} to 1$ as $x to 0$, then $sin x$ becomes very close to $x$.)
In case of your limit, you have
$$lim_{m to infty} frac{sinleft(frac{pi}{2^m}right)}{frac{pi}{2^m}}$$
It’s clear that as $m to infty$, $frac{pi}{2^m} to 0$ due to the growing denominator, so we make a substitution $t = frac{pi}{2^m}$. Clearly, $t to 0$ as $m to infty$, so the limit becomes
$$lim_{t to 0}frac{sin t}{t}$$
which becomes $1$.
answered yesterday
KM101
5,3691423
add a comment |
Because we have
$$lim_{x to 0}frac{sin x}{x} = 1$$
which is a standard limit typically proven through simple geometrical observations followed by the Squeeze Theorem. You can check this link for a various explanation of why this works, particularly the one by robjohn, which I find the simplest and most elegant.
Another way of stating the above limit is
$$sin x sim x; quad x to 0$$
(Obviously, if $frac{sin x}{x} to 1$ as $x to 0$, then $sin x$ becomes very close to $x$.)
In case of your limit, you have
$$lim_{m to infty} frac{sinleft(frac{pi}{2^m}right)}{frac{pi}{2^m}}$$
It’s clear that as $m to infty$, $frac{pi}{2^m} to 0$ due to the growing denominator, so we make a substitution $t = frac{pi}{2^m}$. Clearly, $t to 0$ as $m to infty$, so the limit becomes
$$lim_{t to 0}frac{sin t}{t}$$
which becomes $1$.
answered yesterday
KM101
5,3691423
Because we have
$$lim_{x to 0}frac{sin x}{x} = 1$$
which is a standard limit typically proven through simple geometrical observations followed by the Squeeze Theorem. You can check this link for a various explanation of why this works, particularly the one by robjohn, which I find the simplest and most elegant.
Another way of stating the above limit is
$$sin x sim x; quad x to 0$$
(Obviously, if $frac{sin x}{x} to 1$ as $x to 0$, then $sin x$ becomes very close to $x$.)
In case of your limit, you have
$$lim_{m to infty} frac{sinleft(frac{pi}{2^m}right)}{frac{pi}{2^m}}$$
It’s clear that as $m to infty$, $frac{pi}{2^m} to 0$ due to the growing denominator, so we make a substitution $t = frac{pi}{2^m}$. Clearly, $t to 0$ as $m to infty$, so the limit becomes
$$lim_{t to 0}frac{sin t}{t}$$
which becomes $1$.
answered yesterday
KM101
5,3691423
answered yesterday
KM101
5,3691423
answered yesterday
KM101
5,3691423
answered yesterday
KM101
5,3691423
5,3691423
add a comment |
add a comment |
If you replace $fracpi{2^m}$ with x, it is obvious that $x$ goes to zero when $m$ goes to infinity. So your limit is equivalent to:
$$lim_{x to 0}frac{sin x}{x}$$
...which is 1. If you learn to evaluate just one limit, this is the one :) You can easily find several elementary proofs on the web and I really recommend this one from MSE.
answered yesterday
Oldboy
7,1191832
add a comment |
If you replace $fracpi{2^m}$ with x, it is obvious that $x$ goes to zero when $m$ goes to infinity. So your limit is equivalent to:
$$lim_{x to 0}frac{sin x}{x}$$
...which is 1. If you learn to evaluate just one limit, this is the one :) You can easily find several elementary proofs on the web and I really recommend this one from MSE.
answered yesterday
Oldboy
7,1191832
add a comment |
If you replace $fracpi{2^m}$ with x, it is obvious that $x$ goes to zero when $m$ goes to infinity. So your limit is equivalent to:
$$lim_{x to 0}frac{sin x}{x}$$
...which is 1. If you learn to evaluate just one limit, this is the one :) You can easily find several elementary proofs on the web and I really recommend this one from MSE.
answered yesterday
Oldboy
7,1191832
If you replace $fracpi{2^m}$ with x, it is obvious that $x$ goes to zero when $m$ goes to infinity. So your limit is equivalent to:
$$lim_{x to 0}frac{sin x}{x}$$
...which is 1. If you learn to evaluate just one limit, this is the one :) You can easily find several elementary proofs on the web and I really recommend this one from MSE.
answered yesterday
Oldboy
7,1191832
answered yesterday
Oldboy
7,1191832
answered yesterday
Oldboy
7,1191832
answered yesterday
Oldboy
7,1191832
7,1191832
add a comment |
add a comment |
dfd is a new contributor. Be nice, and check out our Code of Conduct.
dfd is a new contributor. Be nice, and check out our Code of Conduct.
dfd is a new contributor. Be nice, and check out our Code of Conduct.
dfd is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3060629%2fwhy-is-lim-m-to-infty-frac-sin-left-frac-pi2m-right-frac-pi2%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
