I'm trying to understand the notion of uniform convergence, say that I have a squence of functions $(f_n(x)= (1 -\frac{x}{n}\sin(x))^{-n})_{n \in \mathbb{N}}$. We know that it converges to $e^{x\sin{x}}$. How to show from the definition of uniform convergence that it converges uniformly on every bounded interval of $\mathbb{R}.$
We have to show that for an arbitrary bounded interval $I \subset \mathbb{R}$ we have $$\forall_{\varepsilon >0} \exists_{N} \forall_{n \geq N} \forall_{x \in I} |f_n(x)-f(x)| < \varepsilon.$$
So I fix $\varepsilon >0$ and now I have no idea how to continue. I guess that instead of $\sin{x}$ I could write any continuous function and the proof will be same, but I'm not sure about it. It would be great if anyone could help me.
Let $f_n=(1+\frac {xsinx}{n})^n$(it's the same,for easiest computations) eitherway,
let $g_n(x)=f_n(x)-e^{xsinx}$. Compute $g'_n(x)$=http://www.wolframalpha.com/input/?i=%28%28%28n%2Bx+sin%28x%29%29%2Fn%29%5E%28n-1%29-e%5E%28x+sin%28x%29%29%29+%28sin%28x%29%2Bx+cos%28x%29%29%3D0,find $sup|g_n(x)|_{x\in I}$ and see that for a bounded interval $I$ you have that $sup|g_n(x)|_{x\in I}\to 0$.