Is $f_n=\sqrt{x^2+\frac{1}{n}}$ uniformly convergent on $D=(-\infty,+\infty)$?

Question

Is $f_n$ uniformly convergent on $D=(-\infty,+\infty)$ ? $$ f_n=\sqrt{x^2+\frac{1}{n}} $$

My solution $$ \begin{aligned} &f(x)=\lim_{n\to\infty}f_n=|x|\\ &\lim_{n\to\infty}\sup_D|f_n-f|= \sup_D||x|-|x||=0\ \ \ \forall x\in D \end{aligned} $$

So $f_n$ converges uniformly on $D$.

Am I correct?

Answer 1

Your argument is not valid since you cannot take the supremum after taking the limit.

$|f_n(x)-f(x)|=\frac {(x^{2}+\frac 1n ) -x^{2}} {\sqrt {x^{2}+\frac 1n } +|x|} \leq \frac {1/n} {\frac 1 {\sqrt n}}=\frac 1 {\sqrt n}$ and this prove uniform convergence.