Convergence of $2x^{2n-1}(1+x^{2n})^{\frac{1-n}{n}}$ I have to calculate the limit 
$\lim_{n\to +\infty} 2x^{2n-1}(1+x^{2n})^{\frac{1-n}{n}}$
and to find out if the sequence of function converges uniformly.
I foud that the limit is $0$ if $|x|<1$ and $\pm\infty$ if $|x|>1$, but I'm not sure if I'm right.
Then I have to study if $f_n (x) = 2x^{2n-1}(1+x^{2n})^{\frac{1-n}{n}}$ converges uniformly. I tried with some estimations, but I can't find a $sup$ that converges if $n\to + \infty$
Thank you in advice
 A: Since the functions are odd, we can first assume $x \geq 0$.
$$\frac{2x^{2n-1}}{(1+x^{2n})^\frac{n-1}{n}} = \frac{2x^{2n-1}}{(1+x^{2n})^{\left(1-\frac{1}{n}\right)}} = \frac{2x^{2n}\frac{(1+x^{2n})^\frac{1}{n}}{x}}{1+x^{2n}} = \frac{2x^{2n}}{1+x^{2n}}\left(x^n + \frac{1}{x^n}\right)^\frac{1}{n}$$
If $x > 1$ then $\frac{1}{x^n} \le x^n$:
$$x = \sqrt[n]{x^n} \le \left(x^n + \frac{1}{x^n}\right)^\frac{1}{n} \leq (2x^n)^\frac{1}{n} = \sqrt[n]{2}\cdot x\xrightarrow{n\to\infty} x$$
$$\frac{2x^{2n}}{1+x^{2n}} = \frac{2}{\frac{1}{x^{2n}} + 1}\xrightarrow{n\to\infty} 2$$
So:
$$\lim_{n\to\infty} \frac{2x^{2n-1}}{(1+x^{2n})^\frac{n-1}{n}} = 2x$$
If $x = 1$:
$$\lim_{n\to\infty} \frac{2x^{2n-1}}{(1+x^{2n})^\frac{n-1}{n}} = 1$$
If $x < 1$, we have $\lim_{n\to\infty} 2x^{2n-1} = 0$ and $\lim_{n\to\infty} (1+x^{2n})^\frac{n-1}{n} = 1$. Thus:
$$\lim_{n\to\infty} \frac{2x^{2n-1}}{(1+x^{2n})^\frac{n-1}{n}} = 0$$
We can conclude: 
$$\lim_{n\to\infty} \frac{2x^{2n-1}}{(1+x^{2n})^\frac{n-1}{n}} = \begin{cases}
0,  & \text{if $x < 1$} \\
1,  & \text{if $x = 1$} \\
2x,  & \text{if $x > 1$} \\
\end{cases}$$
Expanding the domain to $\mathbb{R}$, we get:
$$\lim_{n\to\infty} \frac{2x^{2n-1}}{(1+x^{2n})^\frac{n-1}{n}} = \begin{cases}
2x,  & \text{if $x < -1$} \\
-1,  & \text{if $x = -1$} \\
0,  & \text{if $|x| < 1$} \\
1,  & \text{if $x = 1$} \\
2x,  & \text{if $x > 1$} \\
\end{cases}$$
Functions $x \mapsto \frac{2x^{2n-1}}{(1+x^{2n})^\frac{n-1}{n}}$ are continuous for every $n \in \mathbb{N}$. Since the limit function is not continuous, the convergence cannot be uniform.
Edit:
Let's check if the convergence is uniform in $[0,a]$ for $a < 1$:
Let $f_n(x) = 2x^{2n-1}(1+x^{2n})^{\frac{1-n}{n}}$. The derivative is:
$$f_n'(x) = -4(n-1)x^{4n-2}(1+x^{2n})^{\left(-1+\frac{1-n}{n}\right)} + 2(2n-1)x^{2n-2}(1+x^{2n})^\frac{1-n}{n}$$
Rearranging $f'_n(x) = 0$ gives:
$$\frac{x^{2n}}{1+x^{2n}} = \frac{1}{2}\frac{2n-1}{n-1}\implies x^{2n}=1-2n < 0$$
Thus, $f_n$ has no stationary points. This implies that the supremum on $[0,a]$ is $|f(a)| = 2a^{2n-1}(1+a^{2n})^{\frac{1-n}{n}}$.
$$\|f_n - 0\|_{\infty,[0,a]} = \|f_n\|_{\infty,[0,a]} = 2a^{2n-1}(1+a^{2n})^{\frac{1-n}{n}} \xrightarrow{n\to\infty} 0$$
Thus, $f_n \xrightarrow{n\to\infty} 0$ uniformly on $[0,a]$.
