uniform convergence of sequence of function

Question

I have a sequence of function $f_n$:

$$ f_n(x) = \sqrt{x^2 + \frac1n} \qquad \text{on the interval } [-1,1] $$

and $$f(x) = |x| $$

I need to prove that the sequence of functions $f_n$ is uniformly convergent to $f$ on the interval [−1, 1].

$f_n$ is continuously differentiable whereas the limit $f$ is not differentiable at $x = 0$.

Doesn't it contradict the theorem about the differentiability of the limit?

Answer 1

For the uniform convergence notice that $f_n(x)\geq|x|$, but: $$ f_n(x)-|x|=\frac{1}{n(f_n(x)+|x|)}\leq\frac{1}{\sqrt{n}},$$ with equality just in the origin.

Answer 2

We have $f_n(x)>f(x)$ for all $n$ as the square root function is monotonically increasing. Recall that $\sqrt{a+b} \le \sqrt a + \sqrt b$ (this is easily verified). Then $$ f_n(x) = \sqrt{x^2 + \frac1n} \leqslant \sqrt{x^2}+\sqrt{\frac1n} = |x| + \sqrt{\frac1n},$$ so that for any $x$, $$ |f_n(x)-f(x)|=f_n(x)-f(x) \leqslant |x| + \sqrt{\frac1n} - |x| = \sqrt{\frac1n}. $$ From this it is evident that $f_n$ converges uniformly to $f$.