On uniform convergence of $f_n(x) = (2^n +|x|^n)^{1/n}$ for $x\in \mathbb{R}$ Let  $$f_n(x) = (2^n +|x|^n)^\frac{1}{n}, ~~~x\in \mathbb{R}$$

My question is: Does the sequence of ${f_n}$ converge uniformly on $\mathbb{R}$?

I was thinking about $M_n$ test but could not be able to find it out?
Any hint or solution would be appreciated.

Note first that the function defined by $\frac{\exp(u)-1}{u}$ for $u\in ]0,1]$ and $1$ at $0$ is continuous on $[0,1]$, hence bounded, say by $c>0$ and that for $x\geq 0$ we have $\log(1+x)\leq x$.
Put on $[0,1]$, $g_n(v)=(1+v^n)^{1/n}=\exp(\frac{\log (1+v^n)}{n})$. Applying the above inequalities, we find that for $v\in [0,1]$ $$|g_n(v)-1|\leq c\frac{v^n}{n}$$
Now, for $|x|\leq 2$, we have $f_n(x)=2g_n(\frac{|x|}{2})$, hence 
$$|f_n(x)-2|\leq 2 c\frac{|x|^n}{n2^n}\leq \frac{2c}{n}$$
For $|x|\geq 2$, $f_n(x)=|x|g_n(\frac{2}{|x|})$
$$|f_n(x)-|x||\leq |x| c\frac{2^n}{n|x|^n}\leq 2c\frac{2^{n-1}}{n|x|^{n-1}}\leq\frac{2c}{n}$$
Hence 
$$|f_n(x)-\max\{2,|x|\}|\leq \frac{2c}{n}$$
for all $x\in \mathbb{R}$, and the convergence is uniform.
