I have to study the pointwise convergence and uniform convergence of the sequence of functions $$f_n(x)=\sqrt[n]{1+x^{2n}}, \hspace{4mm}x\in \mathbb{R}$$ It is easy to see that
$$
f(x)=
\begin{cases}
1,&\text{if}\, |x|\leq 1\\
x^2,&\text{if}\, |x|>1
\end{cases}
$$
Since $f(x)$ is continous, we still cannot say anything about the uniform convergence of $f_n(x)$ $($remember that continuity of $f(x)$ is a necessary condition for the uniform convergence of $f_n(x)$$)$.
Now, we have to study separately $$sup_{|x|\leq 1}|f_n(x)-1|\hspace{8mm}(1)$$
and $$sup_{|x|>1}|f_n(x)-x^2|\hspace{8mm}(2)$$
and see if they both tend to zero as $n$ goes to infinity.
We first start with $(1)$. So, by the mean value theorem we get $$|\sqrt[n]{1+x^{2n}}-1|=|\frac{1}{n}(1+x^{2n})^{\frac{1}{n}-1}\cdot 2nx^{2n-1}||x-0|=\frac{2nx^{2n}}{n(1+x^{2n})^{1-\frac{1}{n}}}\leq 2\neq 0$$
So does this prove that $f_n(x)$ is not uniformly convergent? I think I have commited an error somehow. Any help would be really appreciated.
Best Answer
The Problem
It is not the Intermediate Value Theorem, which has nothing to do with derivatives, that you are citing, but the Mean Value Theorem, and that is being misquoted, as well.
Applying the Mean Value Theorem
The Mean Value Theorem says that if $f$ is continuous on $[0,x]$, and differentiable on $(0,x)$, that $$ \frac{f(x)-f(0)}{x-0}=f'(\xi)\tag1 $$ for some $\xi\in(0,x)$.
Therefore, for some $\xi\in(0,x)$, $$ \frac{\sqrt[n]{1+x^{2n}}-1}{x-0}=\frac1n\left(1+\xi^{2n}\right)^{\frac1n-1}2n\xi^{2n-1}\tag2 $$ Thus, for $x\lt1$, as $n\to\infty$, $$ \begin{align} \sqrt[n]{1+x^{2n}}-1 &=\overbrace{\quad2x\quad\vphantom{{\xi^2}^{\frac1n}}}^{ \lt2}\overbrace{\ \ \ \xi^{2n-1}\ \ \ \vphantom{{\xi^2}^{\frac1n}}}^{\to0}\overbrace{\left(1+\xi^{2n}\right)^{\frac1n-1}}^{\le1}\tag{3a}\\[6pt] &\le2x^{2n}\tag{3b} \end{align} $$ That is, for $|x|\lt1$, $$ \lim_{n\to\infty}\left(\sqrt[n]{1+x^{2n}}-1\right)=0\tag4 $$
Applying Bernoulli
Simpler is to apply Bernoulli's Inequality, which says that for $n\ge1$ and $|x|\le1$, $$ \begin{align} \sqrt[n]{1+x^{2n}}-1 &\le\frac1nx^{2n}\tag{5a}\\ &\le\frac1n\tag{5b} \end{align} $$ This also gives $(4)$.
The case for $x\gt1$ can be handled as in this answer.