Showing that $f_n(x)=\left(x^n+e^{-nx}\right)^{1/n}$ converges uniformly to $f(x)=\max(x,e^{-x})$

Question

$f_n\to f$ uniformly on $[0,+\infty)$. I've proved pointwise convergence, but I'm struggling with the uniform one. I've tried both through the definition and through Cauchy's condition: I couldn't really go anywhere with Cauchy, while with the definition I tried studying the sign of the derivatives of the differences, but it got too complicated

Answer 1

$f_n(x)$ is equal to $e^{-x}(1+(xe^x)^n)^{1/n}$ when $xe^x\le1$ and equal to $x(1+(\frac{1}{xe^x})^n)^{1/n}$ when $xe^x\ge1$. Thus we only need to show uniform convergence for the following function for the range $0<y\le1$: $$g_n(y):=(1+y^n)^{1/n}$$ where $y$ corresponds to $xe^x$. For $y\ge1$, substitute $1/y$ instead of $y$.

That $g_n(y)\to1$ uniformly is obvious since $1\le g_n(y)\le2^{1/n}$ independently of $y$.