Sin City

Calculus Level 2

lim n sin ( sin ( sin ( ( sin ( x ) ) ) ) n number of sines \large \lim_{n\to\infty} \underbrace{\sin( \sin ( \sin (\cdots ( \sin(x)))\cdots )}_{n \text{ number of sines}}

Compute the limit above, where x x is any real number .

1 2 \frac12 0 0 3 2 \frac{\sqrt3}2 1 2 \frac1{\sqrt2}

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3 solutions

Sabhrant Sachan
May 29, 2016

f ( x ) = sin x [ 1 , 1 ] f ( x ) = sin ( sin x ) [ sin ( 1 ) , sin ( 1 ) ] f ( x ) = sin ( sin ( sin x ) ) [ sin ( sin ( 1 ) ) , sin ( sin ( 1 ) ) ] f(x)=\sin{x} \in [-1,1] \\ f(x)=\sin{\left(\sin{x}\right)} \in [\sin{(-1)},\sin{(1)}] \\ f(x)=\sin{\left(\sin{\left(\sin{x}\right)}\right)} \in [\sin{\left(\sin{(-1)}\right)},\sin{\left(\sin{(1)}\right)}]

As you can see,the Range of the function is deceasing , when the lim n \displaystyle\lim_{n \to \infty} , Range ( R ) 0 (R) \rightarrow 0 , Look at this graph for better understanding.

Yep. You can see that it does approach to y=0. graph graph

Micah Wood - 5 years ago
Keith Rogers
Jun 13, 2016

If the limit exists, call it L. Then the problem is when is L = sin (L). That is only true at y = 0.

You need to prove that the limit exists first.

Pi Han Goh - 5 years ago

It is given for any real number i put x=0.

This is not a solution/proof.

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It is just a value of x which satisfy the given equation and is rational.

Rishabh Deep Singh - 5 years ago

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