<s>[INST] The function $\lim_{n\to\infty}({4^n+x^{2n}+\frac{1}{x^{2n}})}^{1/n}$ is non-derivable at 
The function $$f(x)=\lim_{n\to\infty}\left(4^n+x^{2n}+\frac{1}{x^{2n}}\right)^{\frac{1}{n}}$$ is non-derivable at how many points?

The limit is of $\infty^0$ form. Is it an indeterminate form or is it simply equal to $1$? 
$$\lim_{n\to\infty}\left(4^n+x^{2n}+\frac{1}{x^{2n}}\right)^{\frac{1}{n}}=\lim_{n\to\infty}\frac{(4^nx^{2n}+x^{4n}+1)^{\frac{1}{n}}}{x^2}=\frac{1}{x^2}$$
 [/INST] If $\frac14 < x^2 < 4$, then we can write
$$\begin{align}
\lim_{n\to \infty}\left(4^n+x^{2n}+x^{-2n}\right)^{1/n}&=\lim_{n\to \infty}4\left(1+\left(\frac{x^2}{4}\right)^n+\left(\frac{x^{-2}}{4}\right)^n\right)^{1/n}\\\\
&=4
\end{align}$$
If $x^2>4$, then
$$\begin{align}
\lim_{n\to \infty}\left(4^n+x^{2n}+x^{-2n}\right)^{1/n}&=\lim_{n\to \infty}x^2\left(1+\left(\frac{4}{x^2}\right)^n+\left(\frac{1}{x^4}\right)^n\right)^{1/n}\\\\
&=x^2
\end{align}$$
If If $x^2<1/4$, then
$$\begin{align}
\lim_{n\to \infty}\left(4^n+x^{2n}+x^{-2n}\right)^{1/n}&=\lim_{n\to \infty}x^{-2}\left(1+\left(4x^2\right)^n+\left(x^4\right)^n\right)^{1/n}\\\\
&=x^{-2}
\end{align}$$
Therefore, we can write
$$f(x)=\begin{cases}x^2&,|x|>2\\\\4&\frac12\le |x|\le 2\\\\\frac{1}{x^2}&,0<|x|<\frac12\end{cases}$$
Obviously, $f$ is differentiable everywhere it is defined except when $x\ne \pm \frac12$ and $x=\pm 2$.  Note that $f$ is undefined at $x=0$.
</s>