Proving the uniform convergence of $f_n$

Question

I need to show that $f_n(x) = \frac{\ln(1+nx)}{n+x}$ converges uniformly on $X = [0,10]$. I have tried proving by Cauchy criterion, as well as from the definition (where limit function is $f =0$), but no results so far.

Answer 1

Hint. Let $R>0$. Then for $x\in [0,R]$ and for $n\geq 1$, $$0\leq \frac{\ln(1+nx)}{n+x}\leq \frac{\ln(1+nR)}{n}.$$ What may we conclude as $n$ goes to infinity?

P.S. Actually, the convergence is uniform in $[0,+\infty)$. Since $\ln(1+nx)$ is concave, its graph stays under the tangent line at $x=n$: $$\ln(1+nx)\le \ln(1+n^2)+\frac{n(x-n)}{1+n^2}.$$ Therefore, for $x\geq 0$, $$0\leq \frac{\ln(1+nx)}{n+x}\leq \frac{\ln(1+n^2)}{n+x}+\frac{n}{1+n^2}\cdot \frac{x-n}{n+x}\leq \frac{\ln(1+n^2)}{n}+\frac{n}{1+n^2}.$$