Determine pointwise and uniform convergence of a given sequence of functions and calculate an integral

Question

Let $f_n:[0,\infty)\to\mathbb{R},f_n(x)=\frac {ne^{-x}+xe^{-n}}{n+x},\space\space\space\forall n\in\mathbb{N}$. Study the convergence and calculate: $A_n=\int_0^1f_n(x)dx.$

My attempt:

For the pointwise convergence we take the limit to $\infty$ of $f_n$. And we have:

$$\lim_{n\to\infty} \frac {ne^{-x}+xe^{-n}}{n+x}=1,\forall\space\space\space x\geq0.$$

For the uniform convergence we have to calculate:

$$\lim_{n\to\infty}\sup_{x\geq0}\left|\frac {ne^{-x}+xe^{-n}}{n+x} - 1\right|.$$

I took $g:[0,\infty)\to\mathbb{R}, g(x)=\frac {ne^{-x}+xe^{-n}}{n+x} - 1$

But if we take the derivative it gets too complicated, what other ways are there to determine the uniform convergence?

Answer 1

Hint. Actually $$\lim_{n\to\infty} \frac{ne^{-x}+xe^{-n}}{n+x}=e^{-x}.$$ Hence for the uniform convergence you should consider the limit as $n\to \infty$ of the left-hand side $$\sup_{x\geq 0}\frac{x|e^{-x}-e^{-n}|}{n+x}\leq \sup_{x\geq 0}\frac{xe^{-x}}{n+x} +\sup_{x\geq 0}\frac{xe^{-n}}{n+x}.$$

Answer 2

First verify that the $\forall n\in\Bbb N\ \sup\limits_{x\ge0} \frac{ne^{-x}+xe^{-n}}{n+x}$ is attained when $x$ goes to zero from the right.

And then compute that $\sup$

It's 1