Purely out of interest, I wanted to try and construct a sequence of differentiable functions converging to a non-differentiable function. I began with the first non-differentiable function that sprung to my mind, namely
\begin{align}
&f:\mathbb{R}\to\mathbb{R}\\
&f(x)=|x|.
\end{align}
After some testing I considered the function defined by
$$f_\varepsilon(x) = |x|+\frac{\varepsilon}{|x|+\sqrt{\varepsilon}} $$
for some $\varepsilon>0$. Then $\lim\limits_{\varepsilon\to0^+}f_\varepsilon(x)=f(x)$, and $f_\varepsilon(x)$ looks smooth, i.e. differentiable for every $\varepsilon>0$ on the entire domain.
Question: How can I prove that $f_\varepsilon$ is differentiable for every $\varepsilon>0$ (or disprove) using the definition of the derivative?
If this assertion is true, then I construct the sequence simply by setting $\varepsilon = 1/n$ for $n\in\mathbb{N}$.
Attempt: I set up the definition for the derivative
\begin{align}
\frac{\mathrm{d}f_\varepsilon}{\mathrm{d}x} &= \lim_{h\to 0}\frac{1}{h}\left[\left(|x+h|+\frac{\varepsilon}{|x+h|+\sqrt{\varepsilon}}\right)-\left(|x|+\frac{\varepsilon}{|x|+\sqrt{\varepsilon}}\right)\right]\\
&=\lim_{h\to 0}\frac{1}{h}\left[|x+h|-|x|+\frac{\varepsilon}{|x+h|+\sqrt{\varepsilon}}-\frac{\varepsilon}{|x|+\sqrt{\varepsilon}}\right],
\end{align}
but I could not figure out how to proceed.
Sidenotes: An interesting thing I discovered when constructing $f_\varepsilon$, was that almost any small change removes its smoothness, for example
\begin{equation}
g_\varepsilon(x) = |x|+\frac{2\varepsilon}{|x|+\sqrt{\varepsilon}}\hspace{2cm} h_\varepsilon(x) = |x|+\frac{\varepsilon}{|x|+2\sqrt{\varepsilon}}
\end{equation}
do both not look smooth at all. Similarly for the other terms; changing the coefficients will remove the smoothness. I am also somewhat intrigued by this. So if anyone can shed some light on this, even better.
Best Answer
Consider the sequence of differentiable functions $h_n(x) = x^{1+\frac{1}{2n-1}}$ defined on $[-1,1]$ and note $$ \lim_{n\rightarrow\infty} h_n(x) = x\lim_{n\rightarrow\infty} x^\frac{1}{2n-1}= |x|,\qquad \forall x\in[-1,1]. $$