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

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.
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  $\lim\limits_{n\rightarrow \infty} \sup\limits_{x\in (-1,1)\setminus \{0\}} |f_n(x)-1|&gt;0$
Since $$\left|\frac{\sin u}{u} -1\right|=\left|\frac{\sin u -u}{u}\right|\leqslant 1- \cos u$$ we have $$\left|\frac{\sin (x^n )}{x^n} -1\right|=\left|\frac{\sin (x^n ) -x^n}{x^n}\right|\leqslant 1- \cos (x^n )$$

and since the sequence $(x^n ) $ tends to $0$ almost uniformly on $(-1,1 )$ thus the sequence $\left( \frac{\sin (x^n )}{x^n} \right)$ tends to $0$ almost uniformly on $(-1,1 ).$