Proving uniform convergence using MVT for definite integrals

Question

Let $(e^{x/n})$ be a sequence of functions where $x \in [0,a]$ (a is a positive real number). I need to prove that this converges uniformly to 1 using the MVT for integrals.

Here's what I've done so far:

$|f_n(x)-f(x)|=|e^{x/n}-1|=|\int_0^{x/n} \! e^t \, \mathrm{d}t|$. Now since $e^{x/n}$ is continuous, by the MVT there exists a number $c \in (0,x/n)$ so that $\int_0^{x/n} \! e^t \, \mathrm{d}t=e^cx/n$. Thus, $|\int_0^{x/n} \! e^t \, \mathrm{d}t|=|e^cx/n|\leq|e^{x/n}x/n|\leq|e^{a/n}a/n|$.

To finish the proof, I want to find an N<n for which $|e^{a/n}a/n|<\epsilon$, but it seems impossible to do so. What am I doing wrong?

Answer 1

Observe that $e^{a/n}\le e^a$ for all $n\ge 1$ and then $|e^{an}a/n|\le e^a/n$. So all you need is $n\ge N:= e^a/\varepsilon +1$.

Answer 2

I don't see the motivation for the use of MVT for integrals when this follows immediately from the definition of uniform convergence: $$ \sup_{x\in[0,a]}|e^{x/n}-1|=|e^{a/n}-1|\stackrel{n\to\infty}\longrightarrow0. $$