Limit of infinite sines

[talolard]

Hello,

Homework Statement

Calculate the limit of $$lim_{n-> /infty} sinsinsin..._{n times } sin(x)$$ where x is any real number


My attempt:

Using the Taylor expansion we know that $$sin(x)=x- \frac{x^3}{3!} +\circ(x^3)$$ and that $$1 \geq sin(x) \geq -1$$
If the expression has a limit then every subsequence aproaches the same limit. specifically we can define a sequence
$$a_{n+1}=|a_{n}- \frac {a_{n}^3}{3!}|$$ and so the limit of this sequence must be the limit of the above function.
We can restrict our inquiry to $$a_1\in[-1,1]$$ since tex]1 \geq sin(x) \geq -1 [/tex]

We will show by induction that the sequence is montonic decreasing:
$$a_{2}=|a_{1}- \frac {a_{1}^3}{3!}| \geq |a_{1}| - |\frac {a_{1}^3}{3!}| /leq a_{1}$$ which holds for any value of $$a_1$$
We will assume for n and prove for n+1
$$a_{n+1}=|a_{n}- \frac {a_{n}^3}{3!}| \leq^{?} a_{n} \iff | \frac {3!a_{n} - a_{n}^3}{3!} < a_{n}| \iff |3!a_{n} - a_{n}^3| \leq 3!a_{n} -> |3!a_{n} - a_{n}^3| \geq |3!a_{n}| -| a_{n}^3| \leq 3!a_{n} \iff -| a_{n}^3| \leq 0$$
Or in words, we used the traingle inequality and the fact that the minimum of an absolut value is zero to show that the sequence decreases.

There is no need to show that it is bounded since an absolute value is lowerly bounded by 0 and we have shown that the sequence is monotonically decreasing.
So we knwo the sequence converges to a finite limit and so the original function does as well. but how do i show that the limit is zero?

Thanks for the help,
Tal

 

[mighty2000]

Hello Tal,

Your approach is definitely on the right track. To show that the limit is zero, we can use the squeeze theorem. We know that the limit of a_{n+1} is equal to the limit of the original function, so if we can show that a_{n+1} approaches 0, then we can conclude that the original function also approaches 0.

To do this, we can use the fact that 1 \geq sin(x) \geq -1 and the Taylor expansion of sin(x) to show that a_{n+1} is always less than or equal to a_{n}. This is because the term \frac{a_n^3}{3!} will always be positive, so subtracting it from a_n will make a_{n+1} smaller than a_n. This means that a_{n+1} is always between -a_n and a_n, and since we know that a_n approaches 0, we can conclude that a_{n+1} also approaches 0. Therefore, the limit of the original function is also 0.

I hope this helps clarify things. Keep up the good work with your mathematical reasoning!