I am stuck on determining whether this sequence of functions converges uniformly or not.
Let $f_n(x)=\frac{1}{n}\log(\frac{1+x}{n})$ on the interval $\left(0,+\infty\right)$. I can prove that $\lim_{n\to +\infty} f_n(x)=0$ which means that the sequence $f_n$ converges pointwise to the function $f(x)=0$.
Computing $$\sup_{x\in\left(0,+\infty\right)}|f_n(x)-f(x)|=\frac{1}{n}\sup_{x\in\left(0,+\infty\right)}|\log(\frac{1+x}{n})|,$$ let $h_n(x)=\log(\frac{1+x}{n})$ so the derivative fixing $n$ is $\frac{1}{x+1}>0$ for $x\in\left(0,+\infty\right)$.
Is it enough to say that this sequence of functions converges uniformly or not? Im I totally wrong?
For each $n\in\Bbb N$, $f_n(ne^n-1)=1$. So,$$\sup_{x\in(0,\infty)}\bigl|f_n(x)-f(x)\bigr|\geqslant1,$$and therefore the convergence is not uniform.