calculus - $lim_{xtoinfty} frac{sin x+sin^2x+sin^3x+dotsb}{xsin x}$, infinite series limit

I saw this question yesterday.

$$\lim_{x\to\infty} \frac{\sin x+\sin^2x+\sin^3x+\dotsb}{x\sin x}$$

I claim that the limit is $0$ because it can be written like the following.

$$\lim_{x\to\infty} \left(\frac{1}{x}+\frac{\sin x}{x}+\frac{\sin^2x}{x}+\dotsb\right)$$

Then I say in this expression the numerator is limited between $[-1,1]$, and the denominator for each term goes as much as to infinity. Hence term by term we have 0, and adding these up we have 0 as the answer.

But then some guy says its indeterminate, because

$$\lim_{x\to\infty} \frac{1+\sin x+\sin^2x+\dotsb}{x} = \lim_{x\to\infty} \frac{1-\sin^nx}{(1-\sin x)x}$$

He claims what we have in the denominator is oscillating and we cannot have a limit. Also, he says I'm wrong because an infinite amount of zero does not equal to zero.

I'd like to hear your ideas about the answer.

Answer

Let

$$f(x)=\frac{1+\sin x+\sin^2x+\ldots}x$$

for all $x\in\Bbb R$ for which this is defined. If $x>0$ and $\sin x\ne\pm 1$, then

$$f(x)=\frac{1}{x(1-\sin x)}\;.$$

Let $n$ be any odd positive integer. By taking $x$ sufficiently close to $\frac{n\pi}2$, we can make$f(x)$ arbitrarily large. Thus, there is a strictly increasing sequence $\langle x_n:n\in\Bbb N\rangle$ of positive real numbers such that $\lim\limits_{n\to\infty}f(x_n)=\infty$.

Since there is also such a sequence for which $f(x_n)=0$ for each $n\in\Bbb N$, the limit does not exist.