Let $\forall n \in \mathbb{N^*}$ $$ f_n(x) = \frac{ne^{-x}+x^2}{n+x} $$ I know by simple calculation that it pointwise converge to : $f(x)=e^{-x}, \forall x \in \mathbb{R} \setminus \{-n\}$
Now I want to show the uniform converge on a compact $[a,b]$ with $ a,b\in\mathbb{R}^+$, of that sequence using: $$ \lVert f_n - f \rVert_\infty = \sup_{x \in [a,b]}{\lvert \frac{x^2 - xe^{-x}}{n+x} \rvert} $$ At that point I don't know what to do because the derivative is too complex to be usable and it seem hard to bound the norm with something function of n.
Over $\Bbb R$, it cannot be uniform, since $$ \Vert f_n - f \Vert_\infty = \sup_{\Bbb R} \left\vert \frac {x^2 - x\mathrm e ^{-x}} {n + x} \right\vert \geqslant \frac {n^4 - n^2 \mathrm e^{-n^2}} {n + n^2}, $$ where $$ \left\vert \frac {n^2 \mathrm e^{-n^2}} {n^2 + n} \right\vert \leqslant e^{-n^2} \xrightarrow {n \to +\infty} 0, $$ and then $$ \Vert f_n - f \Vert_\infty = \sup_{\Bbb R} \left\vert \frac {x^2 - x\mathrm e ^{-x}} {n + x} \right\vert \geqslant \frac {n^4 - n^2 \mathrm e^{-n^2}} {n + n^2} \sim \frac {n^4}{n+n^2} \sim n^2 \xrightarrow {n\to +\infty} +\infty. $$ Therefore if $E$ is an infinite interval of the form $[a,+\infty )$, then it cannot converge uniformly.
For a compact subset $K$ in $\Bbb R$, it is always bounded by some $[-M, M]$. If $K \cap (- \Bbb N^*) = \varnothing$, then $f_n$ is defined for every $n$; otherwise we start the sequence from, say, $n = \lfloor M + 2 \rfloor$, then these $f_n$'s are still defined in $K$. Either way, we could bound the numerator $$ x^2 - x\mathrm e^{-x} $$ on $[-M, M]$ by continuity, and this bound is related only to $M$. Let it be $B$. Also on $[-M, M]$, whenever $n \geqslant \lfloor M+2 \rfloor > M+1$, $$n + x \geqslant n - M \geqslant M+1 - M > 0, $$ so $$ \sup_{x \in K} |f_n - f| \leqslant \sup _{x \in [-M, M]} |f_n - f| \leqslant \frac {B} {n - M} \xrightarrow {n \to +\infty} 0. $$ Therefore $(f_n)_{n = \lfloor M+2 \rfloor}^{+\infty} \rightrightarrows f$ on $K$ compact in $\Bbb R$. $\square$