$f_n:[-1,1] \to \Bbb R$ defined by $f_n(x)=\sqrt{x^2+1/n}$ This sequence of functions are uniformly convergent to $f(x)=|x|$. How to show directly that the derivative of this sequence does not converge uniformly?
 A: *

*Indirect solution:
If it was the case, the limit would be a derivable function in 0, which is not. See wikipedia, for example.

*Direct solution:
let us consider the difference
$$0\le f'(x)-f_n'(x) = \frac x{\sqrt{x^2}} - \frac x{\sqrt{x^2+ \frac 1n}}\\=
x\frac {\sqrt{x^2+ \frac 1n} - \sqrt{x^2}}{\sqrt{x^2+ \frac 1n}\sqrt{x^2}}
\\=
\frac 1n\frac 1{\sqrt{x^2+ \frac 1n} + \sqrt{x^2}} 
\frac 1{\sqrt{x^2+ \frac 1n}}
$$
Now let us try to find the $\sup$ on $(0,1)$ of such quantity.
As $$\sup_{x\in (0,1)}\frac 1n   \frac 1{\sqrt{x^2+ \frac 1n} + \sqrt{x^2}}<\infty $$
then$$
\sup_{(0,1)} |f'-f_n'|<\infty\iff
\sup_{x\in (0,1)}\frac 1{\sqrt{x^2+ \frac 1n}}<\infty
$$but $$\sup_{x\in (0,1)}\frac 1{\sqrt{x^2+ \frac 1n}}=\infty$$

A: We first have that
$$f_n'(x)=\frac{x}{\sqrt{x^2+1/n}}$$
Define
$$f^*(x)=\begin{cases}-1& \text{if }x<0\\0&\text{if }x=0\\1&\text{if }x>0\end{cases}$$
We can clearly see that $f_n'\rightarrow f^*$ pointwise.
It is a general theorem that if $g_n\rightarrow g$ uniformly then $g_n\rightarrow g$ pointwise (It is also important that pointwise limits are unique).
To show that $f_n'$ does not converge uniformly, it then suffices to show that $f_n'\not\rightarrow f^*$ uniformly. In other words, we must show that $$\exists\epsilon>0\,\forall N\in\Bbb N\,\exists n\geq N\,\exists x\in[-1,1]\left(|f_n'(x)-f^*(x)|\geq\epsilon\right)$$
This is the difficult tedious part. We could choose any $\epsilon$ with $0<\epsilon<1$. Then given any $N\in\Bbb N$ we could choose $n=N$. With these choices we just need to find a point $x$ close enough to $0$. I'll leave it for you to verify that $x=\sqrt{\frac{(2-\epsilon)(1-\epsilon)^2}{N\epsilon}}$ works.
A: Observe that each $f_n'(x)$ is continuous on $[-1, 1]$ so the uniform limit $g(x)$ should be continuous on $[-1, 1]$. But $g(x) = 1$ if $x > 0$, $-1$ if $x < 0$ and is 0 at $x = 0$ is clearly not continuous at $x = 0$. 
So $g(x)$ is not continuous on $[-1, 1]$, and the convergence is not uniform.
