Uniform convergence of $\sin(x^n)/x^n$

Question

Let $f_n(x)=\sin(x^n)/x^n$. Show that $f_n$ converges uniformly to $1$ in the interval $(-\delta,\delta)- \{0\}$, with $0<\delta<1$.

So far what I've tried is to take the infinite norm of the function in the interval minus $1$ and see if its limit when $n$ tends to $\inf$ is equal to 0.

It looks like the function is strictly positive and less than 1 in the interval, plus it looks crecient in (-1,0) and decrecient in (0,1), though I haven't managed yet to formalize this.

Answer 1

As another answer points out, this is false for your given interval $(-1,1)$. If you assume your interval is $(-\delta,\delta)$ for any $0 < \delta < 1$ then you will get uniform convergence, because $|x^n|$ will be no more than $\delta^n$ which tends to $0$ uniformly for all $x$ in your interval, and as soon as you have uniform convergence of $x^n$ to $0$ you can fairly easily show your sequence of functions converges uniformly to $1$. For your interval $(-1,1)$, $x^n$ does not converge uniformly to $0$ and so the argument fails.

Answer 2

This is false. Let $f_n(x)=\frac{\sin x^n}{x^n}$ and take $x_n=1-\frac{1}{n}$. Then $|f_n(x_n)-1|\rightarrow 1-\frac{\sin (e^{-1})}{e^{-1}}\neq 0$ Therefore $\lim\limits_{n\rightarrow \infty} \sup\limits_{x\in (-1,1)\setminus \{0\}} |f_n(x)-1|>0$