Uniform Convergence of $\frac{n}{x+n}$

Question

I was trying an exercise on uniform convergence of sequence of real-valued functions.

I got stuck in a problem in which I am supposed to prove that sequence defined by $f_n(x)=\frac{(n)}{(x+n)}$ is uniformly convergent on $[0,k]$.

I have found out its point-wise limit to be $1$.

Now I know that for this to be uniformly convergent,for any given $\epsilon>0$,we should get a natural number $m$ such that $|f_n(x)-1|<\epsilon$ for all $n\geq m$ and for all $x$ in $[0,k]$, which on further calculations give us that $n>x(\frac{1}{\epsilon}-1)$. Now I am not able to find a natural number $m$ such that this holds for all $n\geq m$.

Please help.

Answer 1

Note that$$\left|\frac n{x+n}-1\right|=\frac{|x|}{|x+n|}\leqslant\frac kn,$$since $x\leqslant k$ and $|x+n|=x+n\geqslant n$. So, given $\varepsilon>0$, take $N\in\Bbb N$ such that $\frac kN<\varepsilon$. Then, if $n\geqslant N$, $\frac kn<\varepsilon$ and so$$(\forall x\in[0,k])(\forall n\in\Bbb N):n\geqslant N\implies\left|\frac n{x+n}-1\right|<\varepsilon.$$

Answer 2

$\sup_{x \in [0,k]}|f_n(x)-1|= |\frac{x}{x+n}|\leq \frac{k}{n}\to 0$ as $n \to \infty$.