Is it true that $f_n(x)=\left(\frac{n}{n+x}\right)^n$ uniformly converges to $e^{-x}$

Question

Is it true that: $$f_n(x)=\left(\frac{n}{n+x}\right)^n$$ uniformly converges to $e^{-x}$? I have tried to prove it using definition with supremum, but have not succeed. Can you give me a hint, please?

Answer 1

Hint for uniform convergence on $[0,\infty)$: let $\epsilon >0$ and choose $T$ such that $e^{-x} <\epsilon$ for all $x \geq T$. Next choose $m$ such that $(1+\frac T n)^{n} >\frac 1 {\epsilon}$ for $n \geq m$. [This is possible because $(1+\frac T n)^{n} \to e^{T}>1/\epsilon)$. Now you can easily check that $|(\frac n {n+x})^{n}-e^{-x}| <2\epsilon$ for all $x \geq T$ for all $n \geq m$. For uniform convergence on $[0,T]$ use Dini's Theorem.

Answer 2

Do you know that $\lim_{n\to\infty}\left(1+\dfrac{x}{n}\right)^n=\mathrm{e}^x$?

You may then use that $\left(\dfrac{n}{n+x}\right)^n= \left(\dfrac{n+x}{n}\right)^{-n}= \left(\left(1+\dfrac{x}{n}\right)^n\right)^{-1}$. Continuity of the function $g(y) = y^{-1}$ for $y\neq 0$ then gives the result.