# pointwise convergence does not preserve differentiability

In my real analysis lecture, when the lecturer give an example which shows that pointwise convergence does not preserve differentiability, I am not quite understand. The example is as follow:

Suppose $$h_n(x)=x^{1+\frac{1}{2n-1}}$$ on the set $[-1,1]$. Notice that $$\lim_n{h_n(x)}=\lim_{n}{x^{1+\frac{1}{2n-1}}}=x\lim_n{x^{\frac{1}{2n-1}}}$$ This limit, taking the odd roots of $x$, will tend to $1$ if $x>0$ , $-1$ if $x<0$ and $0$ if $x=0$ , which is eauivalent to $|x|$. Hence, we have $\lim_n{x^{1+\frac{1}{2n-1}}}=|x|$, which is not differentiable

Here is my doubt. How do we know that the limit tends to these values? Is there any way to see it?

You should consider that pointwise convergence means that you take a (stable) $x\in [-1.1]$ and regard it as a number, rather than as a variable. Thus, if $x\neq0$ then $x^{\frac{1}{2n-1}}\longrightarrow 0$ , while if $x=0$ then the limit is again $0$ obviously.

This is not an answer, but is another example that might be easier to follow:

Let $\epsilon > 0$ and $f_\epsilon(x) = \begin{cases} -x & x < \epsilon \\ \frac{1}{2 \epsilon} x^2 + \frac{1}{2\epsilon} & |x| \le \epsilon \\ x & x > \epsilon \end{cases}$

This is an approximation to $x \mapsto |x|$ that is continuously differentiable everywhere (the value and derivative of the separate functions match at $x = \pm \epsilon$). It is easy to see that $f_{\frac{1}{n}}(x) \to |x|$ everywhere, $f_{\frac{1}{n}}$ is $C^1$, but $x \mapsto |x|$ is not differentiable at $x=0$.

You must know the following results, which is rather easy to prove: If $f:I\rightarrow\mathbb{R}$ is a continuous function defined in some interval $I\subseteq\mathbb{R}$ and $\left\{x_n\right\}_{n\in\mathbb{N}}$ is a real sequence which converges for some number $x\in I$, then the sequence $\left\{f(x_n)\right\}_{n\in\mathbb{N}}$ converges to $f(x)$, that is, $\lim f(x_n)=f(\lim x_n)$.

For a fixed $x>0$, the function $t\geq 0\mapsto x^t=\exp(t\cdot\log x)\in\mathbb{R}$ is continuous, Hence

$$\lim_{n\rightarrow\infty}x^\frac{1}{2n-1}=x^{\lim_{n\rightarrow\infty}(1/2n-1)}=x^0=1$$

The case $x<0$ is analogous, since $x^\frac{1}{2n-1}=-|x|^\frac{1}{2n-1}=-exp\left(\frac{1}{2n-1}\log|x|\right)$