I tried to do a uniform convergence exercise of this function:
$$f_n(x)=\ln\left(1+\frac{|x|}{n}\right)\frac{nx}{x^2+n}$$
The pointwise limit of this functions is 0, so for the uniform convergence I've studied:
$$\sup_{x\in[-2, 2]}|f_n(x)-f(x)|$$ so:
$$|f_n(x)-f(x)|=\left|\ln\left(1+\frac{|x|}{n}\right)\frac{nx}{x^2+n}\right|=\frac{\left|\ln\left(1+\frac{|x|}{n}\right)nx\right|}{x^2+n}=\frac{\ln\left(1+\frac{|x|}{n}\right)n|x|}{x^2+n}$$
the last one is a even function so i can study the convergence over $x\in[0, 2]$ and i can delete the absolute module so: $$\frac{\ln\left(1+\frac{x}{n}\right)nx}{x^2+n}$$ now $x^2+n\geq n $:
$$\frac{\ln\left(1+\frac{x}{n}\right)nx}{x^2+n} \leq \frac{\ln\left(1+\frac{x}{n}\right)nx}{n}=x\ln\left(1+\frac{x}{n}\right)$$ Now the derivative of $f_n(x)$ is always greater than $0$ for $x\in[0, 2]$ so that $f_n(x)$ is an increasing function which is equal to $0$ in $f_n(0)$ and equal to $2\ln\left(1+\frac{2}{n}\right)$ in $f_n\left(2\right)$, the function assumes its maximum for $x=2$ which, for $n \to \infty$, goes to $0$ so there is uniform convergence for $x\in[-2, 2]$. Are my reasonings right?