Finding the limit of the integral of some uniformly convergant sequence of functions Let $\{f_n\}_{n\in \mathbb N}$ be a sequence of uniformly convergent functions on the interval [0,1]. Find the limit of the following integral: $$\lim_{n\to \infty} \int_0^1nf_n(t)e^{-nt}\,dt$$
Here is my thought process so far. In isolation l'hospital's rule would indicate that  $\lim_{n\to \infty} \frac{n}{e^{nt}}$ would tend to zero, and so I expect that this exponential term will be more dominating.
Normally Lebesgue's Dominated convergence theorem (or at the very least his monotone convergence theorem) would allow us to say that this uniformly convergent sequence of functions would converge to f...however I am guessing the surrounding terms sufficiently mess this up to the point where we no longer have convergence. My guess is the point of this question is to prove that this is the case.
We know that  $\lim_{n\to \infty}\{f_n\}_{n\in \mathbb N}$ is measurable, but If I can show that
$g_n= \frac{n}{e^{nt}}$ is not measurable for $t\in[0,1]$ (because it tends to zero), then can I say this limit does not exist? I'm pretty confused, as is abundantly clear hahahaha.
 A: Let $(f_n)$ be a sequence of continuous functions that converges uniformly to $f$ on $[0,1]$.  By the triangle inequality, we have
$$\left|\int_0^1 n e^{-nt} f_n(t) \, dt - f(0) \right|\leqslant \left|\int_0^1 n e^{-nt} (f_n(t) - f(t)) \, dt \right|+ \left|\int_0^1 n e^{-nt} f(t) \, dt - f(0) \right|$$
For the first term on the RHS, we have
$$\left|\int_0^1 n e^{-nt} (f_n(t) - f(t)) \, dt \right| \leqslant \int_0^1 n e^{-nt}|f_n(t) - f(t)|\, dt$$
As $f_n \to f$ uniformly, for any $\epsilon > 0$ there exists $N_1 \in \mathbb{N}$ such that if $n > N_1$, then $|f_n(t) - f(t)| < \frac{\epsilon}{2}$ for all $t \in [0,1]$, and
$$\left|\int_0^1 n e^{-nt} (f_n(t) - f(t)) \, dt \right| \leqslant \frac{\epsilon}{2} \int_0^1ne^{-nt} \, dt = \frac{\epsilon}{2} (1 - e^{-n}) < \frac{\epsilon}{2} $$
Thus, for all $n > N_1$,
$$\left|\int_0^1 n e^{-nt} f_n(t) \, dt - f(0) \right|\leqslant \frac{\epsilon}{2}+ \left|\int_0^1 n e^{-nt} f(t) \, dt - f(0) \right|$$
Since $f$ is the uniform limit of a sequence of continuous functions it is continuous, and the proof given here shows there exists $N_2$ such that if $n > N_2$
$$\left|\int_0^1 n e^{-nt} f(t) \, dt - f(0) \right| < \frac{\epsilon}{2}$$
Altogether this proves that
$$\lim_{n \to \infty}\int_0^1 n e^{-nt} f_n(t) \, dt = f(0)$$
