# Prove that for real number $r > 0$ the seq of funcs $f_n(x) = \frac{e^x}{1 + n + x^2}$ converges uniformly on $[−r, r]$ to the 0 function.

Prove that for every real number $r > 0$ the sequence of functions $f_n(x) = \frac{e^x}{1 + n + x^2}$ converges uniformly on $[−r, r]$ to the zero function.

I'm not sure if I did this right.

Work:

The definition of uniform convergence for a sequence of functions is:

Let $E$ be a nonempty subset of $\mathbb{R}$. Sequence $f_n:E \to \mathbb{R}$ converges uniformly iff there is $\epsilon > 0$ and $N \in \mathbb{N}$ such that $n \ge N \implies |f_n(x) - f(x)| < \epsilon$ for all $x \in E$.

So, in this case we have $E = [-r, r]$ and basically we need to show that for $n \ge N$ for $N \in \mathbb{N}$ and $\epsilon >0$ that $|f_n(x)| < \epsilon$.

$\lim_{m \to \infty} f_mx = 0$, so clearly $|f_nx| < \epsilon$, which means the sequence $f_n(x)$ converges uniformly to the zero function on $[-r,r]$.

------------

I'm not sure if I tackled it right, or did it even remotely correctly. Did I? How should one actually solve it?

• Well, you haven't done anything. You need to show that for any $\epsilon>0$ there is some $N$ such that if $n \ge N$ and $|x| \le r$ that $|f_n(x)| < \epsilon$. Hint: It is fairly straightforward to find a convenient upper bound $|f_n(x)|$ Feb 15 '15 at 23:31
• Theres a difference between uniform and pointwise convergence. Feb 15 '15 at 23:33

For any $$x ∈ [−r,r]$$, $$e^x \leqslant e^r$$, and let $$N = [e^r/ϵ]$$. Then for any $$x ∈ [−r,r]$$ and $$n>N$$,
$$|f_{n}(x)|\leqslant \dfrac{{e^r}} {{1+n}}<\dfrac{{e^r}} {{n}}<\dfrac{{e^r}} {{N}}<ϵ$$
So $$f_n(x)→0$$ uniformly on $$[−r,r]$$.
$$\sup\bigg|\frac{e^{x}}{1+n+x^2}-0\bigg| \leq \frac{e^{r}}{1+n} < \frac{e^r}{n} < \epsilon.$$