How to show that $f_n(x)=\frac{n}{x}\ln\left(1+\frac{x}{n}\right), x\in E=(0;10)$ converges uniformly? 
The function is $f_n(x)=\frac{n}{x}\ln\left(1+\frac{x}{n}\right)$, $x\in E=(0;10)$. And I have to prove its uniform convergence.  

It quite easy to prove that it converges to $f(x)=1$: $\lim_{n\rightarrow\infty}\frac{n}{x}\ln\left(1+\frac{x}{n}\right)=\lim_{n\rightarrow\infty}\frac{n}{x}\cdot\frac{x}{n}=1$.
However, it is harder to show that $\left|\frac{n}{x}\ln\left(1+\frac{x}{n}\right)-1\right|\leqslant a_n: \lim_{n\rightarrow\infty}a_n=0$
I understant that $0<f_n(x)<1$ so the last inequality can be rewritten as
$\frac{n}{x}\ln\left(1+\frac{x}{n}\right)\geqslant 1-a_n.$
But that does not help much. 
 A: $\frac {\log (1+t)} t \to 1$  as $t \to 0$. Given $\epsilon >0$ there exists $\delta >0$ such that $|t| <\delta$ implies $|\frac {\log (1+t)} t- 1| <\epsilon$. Now $\frac  1 n  <\frac {\delta} {10}$ (or $n >\frac {10} {\delta}$) implies that $|\frac x n|<\delta$ so   $|\frac n x {\log (1+\frac x n)} - 1| <\epsilon$ for all $x \in (0,10)$.
Finding a $\delta$ explicitly in terms of $\epsilon$: we can show that $t-\frac {t^{2}} 2 \leq \log (1+t) \leq t$ for all $t \geq 0$. Hence $-\frac t 2 \leq \frac {\log (1+t)} t-1 \leq 0$  Hence we can take $\delta=2\epsilon$. 
A: It is not hard to prove that$$x\geqslant0\implies\frac x{x+1}\leqslant\log(1+x).\tag1$$Now, if $f_n(x)=\frac nx\log\left(1+\frac xn\right)$, then$$f_n'(x)=\frac{n\left(x-(n+x)\log\left(1+\frac xn\right)\right)}{x^2(n+x)}$$and (assuming that $x>0$) $f_n'(x)<0$ if and only$$x-(n+x)\log\left(1+\frac xn\right)<0.$$But this assertion is equivalent to$$\frac{\frac xn}{1+\frac xn}<\log\left(1+\frac xn\right),$$which is true, because of $(1)$. So, each $f_n$ is strictly decreasing. Therefore$$(\forall n\in\mathbb N)\bigl(\forall x\in(0,10)\bigr):\bigl\lvert f_n(x)-1\bigr\rvert\leqslant\bigl\lvert f_n(10)-1\bigr\rvert$$and you know that$$\lim_{n\to\infty}f_n(10)-1=0.$$
