Find the following limit: $ \lim_{n \to \infty} \frac{1}{n} \int_0^\infty e^{-x/n}f(x) \, dx. $ The Problem:
Suppose $f$ is Riemann integrable on the interval $[0,A]$, for all $A > 0$ and $f(x) \to -1$ as $x \to \infty$. Find the following limit:
$$ \lim_{n \to \infty} \frac{1}{n} \int_0^\infty e^{-x/n}f(x)\,dx. $$
What I've Tried:
We know that there exist real numbers $m$ and $M$ such that $m \le f(x) \le M$, for all $x \in [0,A]$ (since $f$ is Riemann integrable on the compact interval $[0,A]$). So, we can say that
$$ \frac{m}{n} \int_0^A e^{-x/n} \, dx \le \frac{1}{n} \int_0^A e^{-x/n}f(x) \, dx \le \frac{M}{n} \int_0^A e^{-x/n}\,dx, $$
which implies that
$$ m(1-e^{-A/n}) \le \frac{1}{n} \int_0^A e^{-x/n}f(x) \, dx \le  M(1-e^{-A/n}). $$
So, as $A \to \infty$,
$$ m \le \frac{1}{n} \int_0^\infty e^{-x/n}f(x) \, dx \le  M, $$
which... isn't very helpful.
I feel like the fact that 
$$ \frac{1}{n} \int_0^\infty e^{-x/n} \, dx = 1, $$
for all $n$ (where $n$ is a positive integer),
should be of some use here, but I don't see how. Moreover, I haven't used the fact that $f(x) \to -1$ as $x \to \infty$, so I know I'm not going about this the right way.
I've tried integrating by parts (which, technically, I don't know if I can do since I don't know that $f$ is differentiable); I've tried interchanging limits, etc. Nothing seems to work here...
Also, note that all of the similar problems posted here on MSE usually either involve proper integrals or $f$ is known.
 A: You can prove the integrals are convergent. Let's substitute $t=x/n$, $n\,\text{d}t=\text{d}x$ so we get
$$ \int_0^\infty e^{-t}f(tn)\,\text{d}t $$
$|e^{-t} f(tn)|\leq Me^{-t}$ so that we can use Lebesgue's bounded convergence theorem
$$ \lim_{n\to \infty}\int_0^\infty e^{-t}f(tn)\,\text{d}t = \int_0^\infty \lim_{n\to\infty} e^{-t}f(tn)\,\text{d}t = -\int_0^\infty e^{-t} \, \text{d}t = -1  $$
A: 
I thought it might be instructive to present an approach that does not rely on the theory of Lebesgue integration.  To that end, we proceed.


First, since $\lim_{x\to\infty}f(x)=-1$, then for any $\epsilon>0$, there exists a number $L>0$, such that $|f(x)+1|<\epsilon$ whenever $x>L$.  With a chosen $\epsilon>0$, we fix $L$ such that $|f(x)+1|<\epsilon$.  

Next, we write 
$$\begin{align}
\frac1n\int_0^\infty e^{-x/n}f(x)\,dx&=-1+\frac1n \int_0^\infty e^{-x/n}(f(x)+1)\,dx\\\\
&=-1+\frac1n\int_0^L e^{-x/n}(f(x)+1)\,dx+\frac1n\int_L^\infty e^{-x/n}(f(x)+1)\,dx\tag1
\end{align}$$

The first integral on the right-hand side of $(1)$ is bounded since $f$ is integrable.  Hence, as $n\to \infty$, $\lim_{n\to \infty}  \frac1n\int_0^L e^{-x/n}(f(x)+1)\,dx=0$.

For the second integral, we have
$$\left|\frac1n\int_L^\infty e^{-x/n}(f(x)+1)\,dx\right|\le \epsilon e^{-L/n}<\epsilon$$

We conclude, therefore, that 
$$\lim_{n\to \infty}\frac1n\int_0^\infty e^{-x/n}f(x)\,dx=-1$$
as expected!
