Uniform convergence of $f_n(x) =e^{-x/n} $

Question

I want to check the uniform convergence of $f_n(x) =e^{-x/n} $ with $x>1$ or with $x<1$.

The point wise convergence is 1.

The maximum of the difference does not go to 0 and so we do not have uniform convergence.

Is that correct?

Answer 1

Note that for any $x\in \mathbb R$, $f_n\to 1$.

On $(1,\infty)$,

Note that the $\sup_{x\in (1,\infty)} |f_n(x)-1|=1$. Hence, given any $n$, there is an $x_n\gt 1$ such that $|f_n(x_n)-1|\ge \sup_{x\in (1,\infty)}|f_n(x)-1|-\frac12=\frac 12\gt \frac 14.$ So the convergence $f_n\to 1$ is not uniform on $(1,\infty)$.

The analysis on $(-\infty, -1)$ is simpler as $f_n$'s are unbounded. It follows that for each $n$, there must be some $x_n\lt -1$ such that $f_n(x_n)\gt 5\implies f_n(x_n)-1\gt 4\implies$ the convergence $f_n\to 1$ can't be uniform on $(-\infty, -1)$, which is almost obvious if in the definition of uniform convergence the arbitrary positive number $\epsilon\gt 0$ is chosen as less than $4$.

Answer 2

On $(1,\infty)$ we have $\lim_{x\to \infty}e^{-x/n}=0$ for any $n$, meaning that there will always be values of $x$ where $e^{-x/n}$ is far from its limit. On $(-\infty,1)$ we can always choose an $x$ such that $e^{-x/n}$ is arbitrarily large given fixed $n$, so this the limit is also not uniform on this interval.