I'm attempting to determine if the funtion $$f_n(x) = x^n \sin(\frac{1-x}{x})$$ converges uniformly on (0,1). Obviously, the function converges pointwise to $0$. I've graphed the function and the supremum is strictly less than 1 for n=1 and decreasing for successive n's, leading me to believe that the convergence is indeed uniform. I started typing this out with no idea where to start, but I think I might have an idea now and would like feedback on it. My idea is to show that sequence of functions $f_n(x)$ are equicontinous at $1$.
Let $\epsilon > 0$ be given. It can be shown that $$\lim_{x \to 1} x\sin(\frac{1-x}{x}) = 0$$ so there exists a $\delta > 0$ such that $x \in (1-\delta, 1)$ gives $$\left| x\sin(\frac{1-x}{x}) \right| < \epsilon/2.$$ So, for each $n > 1$ $$\left| x^n \sin(\frac{1-x}{x}) \right| \leq \left|x\sin(\frac{1-x}{x}) \right| < \epsilon$$ for $x \in (1-\delta,1)$. Now for $x \in (0, 1-\delta]$ $$\left| x^n \sin(\frac{1-x}{x}) \right| \leq \left|x^n \right| \leq \left| 1-\delta \right|^n.$$ Since $(1 - \delta) < 1$, we can find $N \in \mathbb{N}$ such that $n \geq N$ gives $$\left| 1- \delta \right|^n < \epsilon.$$ Thus for $n \geq N$, we have $$\left|x^n \sin( \frac{1-x}{x} ) \right| < \epsilon$$ for all $x \in (0,1)$, showing that $f_n(x)$ converges uniformly to $0$.