I would like general feedback on my solution to this exercise, as well as a hint for arriving at the solution more directly. I suspect that my approach is sub-optimal.
Exercise
Does the sequence of functions $ f_n(x) := \sqrt{x^2+\frac{1}{n^2}}$ converge uniformly on $[−1, 1]$? If so, what is the limit function?
Solution
Yes, it converges uniformly to the function $f(x) = |x|$.
$f_n(x)=f_n(-x)$ and $f(x)=f(-x)$ $\forall n,x$ and so by symmetry, it suffices to consider the limit on $[0,1]$ instead.
We have
\begin{align*} \frac{d}{dx}\left(f_n(x)-f(x)\right) &= 0 \\ \frac{x}{\sqrt{x^2+\frac{1}{n^2}}}-1 &= 0 \\ x &= \sqrt{x^2+\frac{1}{n^2}} \\ x^2 &= x^2+\frac{1}{n^2} \end{align*}
which has no solutions, so $f_n(x)-f(x)$ has no stationary points. Since this is a continuous function on a closed interval, its maximum must be at either $x=0$ or $x=1$.
$$ \|f_n-f\|_\text{sup}=\text{max}\left(\sqrt{0+\frac{1}{n^2}}-0, \sqrt{1^2+\frac{1}{n^2}}-1\right)=\text{max}\left(\frac{1}{n}, \sqrt{1+\frac{1}{n^2}}-1\right) $$
Clearly $\frac{1}{n}\to 0$ as $n\to\infty$ and by the algebra of limits $\sqrt{1+\frac{1}{n^2}}-1\to\sqrt{1}-1=0$, so $\|f_n-f\|_\text{sup}\to 0$ and $f_n \to f$ uniformly as required.
$$ \left\lvert \sqrt{x^2 + n^{-2}} - \sqrt{x^2} \right\rvert = \frac{n^{-2}}{\sqrt{x^2 + n^{-2}} + \sqrt{x^2}} \leq \frac{n^{-2}}{n^{-1}} = \frac{1}{n} $$ for any $ x \in \mathbf{R}$; it follows that $\sqrt{x^2 + n^{-2}} \to \lvert x \rvert$ uniformly on $\mathbf{R}$ and hence, in particular, on $[-1, 1]$.