Why is $lim_{mto infty} frac{sinleft(frac{pi}{2^m}right)}{frac{pi}{2^m}} = 1$?
$begingroup$
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 Jan 3 at 19:05
David K
55.5k345120
asked Jan 3 at 15:00
dfddfd
31
$endgroup$
add a comment |
$begingroup$
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 Jan 3 at 19:05
David K
55.5k345120
asked Jan 3 at 15:00
dfddfd
31
$endgroup$
add a comment |
$begingroup$
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 Jan 3 at 19:05
David K
55.5k345120
asked Jan 3 at 15:00
dfddfd
31
$endgroup$
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 Jan 3 at 19:05
David K
55.5k345120
asked Jan 3 at 15:00
dfddfd
31
edited Jan 3 at 19:05
David K
55.5k345120
asked Jan 3 at 15:00
dfddfd
31
edited Jan 3 at 19:05
David K
55.5k345120
edited Jan 3 at 19:05
David K
55.5k345120
edited Jan 3 at 19:05
David K
55.5k345120
55.5k345120
asked Jan 3 at 15:00
dfddfd
31
asked Jan 3 at 15:00
dfddfd
31
asked Jan 3 at 15:00
dfddfd
31
31
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 Jan 3 at 15:16
KM101KM101
6,0801525
$endgroup$
$begingroup$
Thank you so much! :D
$endgroup$
– dfd
Jan 5 at 15:27
add a comment |
$begingroup$
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 Jan 3 at 15:12
OldboyOldboy
9,19611138
$endgroup$
$begingroup$
Thank you so much! :D
$endgroup$
– dfd
Jan 5 at 15:27
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
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 Jan 3 at 15:16
KM101KM101
6,0801525
$endgroup$
$begingroup$
Thank you so much! :D
$endgroup$
– dfd
Jan 5 at 15:27
add a comment |
$begingroup$
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 Jan 3 at 15:16
KM101KM101
6,0801525
$endgroup$
$begingroup$
Thank you so much! :D
$endgroup$
– dfd
Jan 5 at 15:27
add a comment |
$begingroup$
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 Jan 3 at 15:16
KM101KM101
6,0801525
$endgroup$
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 Jan 3 at 15:16
KM101KM101
6,0801525
answered Jan 3 at 15:16
KM101KM101
6,0801525
answered Jan 3 at 15:16
KM101KM101
6,0801525
answered Jan 3 at 15:16
KM101KM101
6,0801525
6,0801525
$begingroup$
Thank you so much! :D
$endgroup$
– dfd
Jan 5 at 15:27
add a comment |
$begingroup$
Thank you so much! :D
$endgroup$
– dfd
Jan 5 at 15:27
$begingroup$
Thank you so much! :D
$endgroup$
– dfd
Jan 5 at 15:27
$begingroup$
Thank you so much! :D
$endgroup$
– dfd
Jan 5 at 15:27
add a comment |
$begingroup$
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 Jan 3 at 15:12
OldboyOldboy
9,19611138
$endgroup$
$begingroup$
Thank you so much! :D
$endgroup$
– dfd
Jan 5 at 15:27
add a comment |
$begingroup$
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 Jan 3 at 15:12
OldboyOldboy
9,19611138
$endgroup$
$begingroup$
Thank you so much! :D
$endgroup$
– dfd
Jan 5 at 15:27
add a comment |
$begingroup$
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 Jan 3 at 15:12
OldboyOldboy
9,19611138
$endgroup$
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 Jan 3 at 15:12
OldboyOldboy
9,19611138
answered Jan 3 at 15:12
OldboyOldboy
9,19611138
answered Jan 3 at 15:12
OldboyOldboy
9,19611138
answered Jan 3 at 15:12
OldboyOldboy
9,19611138
9,19611138
$begingroup$
Thank you so much! :D
$endgroup$
– dfd
Jan 5 at 15:27
add a comment |
$begingroup$
Thank you so much! :D
$endgroup$
– dfd
Jan 5 at 15:27
$begingroup$
Thank you so much! :D
$endgroup$
– dfd
Jan 5 at 15:27
$begingroup$
Thank you so much! :D
$endgroup$
– dfd
Jan 5 at 15:27
add a comment |
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