Dominated Convergence Theorem to Evaluate Integral Assume that $f \in L^1([0,\infty)) \cap C([0,\infty))$. Evaluate the limit $$
\lim_{\epsilon \to 0} \frac{1}{\epsilon}\int_0^\infty f(x) e^{-x/\epsilon} \, dx
$$
Here is what I think: 
First consider $\epsilon>0$. 
(1) It's easy to see that for any $x > 0$, $f(x) \frac{1}{\epsilon}e^{-x/\epsilon} \to 0$ pointwise. 
(2) Using Calculus, it's easy to see that for any $\epsilon>0$, $\frac{1}{\epsilon}e^{-x/\epsilon} \leq \frac{1}{ex}$. Thus, for $x \in (1,\infty)$, $|f(x)\cdot \frac{1}{\epsilon}e^{-x/\epsilon}| \leq \frac{|f(x)|}{ex} \leq \frac{|f(x)|}{e}$, which is integrable. 
(3) For $x \in [0,1]$, $|f(x)|$ is bounded by some constant $C$, so $\displaystyle |f(x)| \frac{1}{\epsilon}e^{-x/\epsilon}\leq C \frac{1}{\epsilon}e^{-x/\epsilon}$, and $\displaystyle \int_0^1 C \frac{1}{\epsilon}e^{-x/\epsilon}\, dx = C \cdot (1-e^{-\epsilon}) \leq C$ for all $\epsilon >0$. 
Hence, by Dominated Convergence Theorem (generalized), we conclude that the integral converges to $0$ as $\epsilon \to 0^+$. 
Then consider $\epsilon <0$, which is where I think the problem lies. 
(4) As $\epsilon \to 0^-$, for each $x \in (0,\infty)$, $f(x)\frac{1}{\epsilon}e^{-x/\epsilon} \to \pm\infty$, so there is no reason why the integral should converge. 
I was wondering if the analysis is correct -- any help is appreciated! 
 A: As you have observed, I think that the limit has to be considered only for $\epsilon \to 0^+$.
Given that, your analysis based on the Dominated Convergence Theorem seems not correct. (I have some doubt for your case (3).)
So, let $\epsilon > 0$ and consider the change of variable $y = x/\epsilon$.
You get
$$
\frac{1}{\epsilon} \int_0^\infty f(x) e^{-x/\epsilon}\, dx
= \int_0^\infty f(\epsilon y) e^{-y}\, dy.
$$
Since $f$ is continous, you can prove that the last integral converges to $f(0)$ as $\epsilon \to 0^+$.
This fact can be proved by splitting the integral in to parts.
Given $K > 0$, we have
$$
\int_0^\infty f(\epsilon y) e^{-y}\, dy =
\int_0^K f(\epsilon y) e^{-y}\, dy
+ \int_K^\infty f(\epsilon y) e^{-y}\, dy.
$$
The second integral at rhs is easily estimated by
$$
\left|\int_K^\infty f(\epsilon y) e^{-y}\, dy\right|
\leq e^{-K} \|f\|_{L^1}.
$$
To the first integral at rhs we can apply the Dominated Convergence Theorem.
Namely, if $0 < \epsilon < 1$ and $M := \max_{[0,K]} |f|$ we get
$$
|f(\epsilon y) e^{-y}| \leq M e^{-y}
\qquad \forall y\in [0, K].
$$
Hence
$$
\lim_{\epsilon\to 0+} \int_0^K f(\epsilon y) e^{-y}\, dy
= f(0) (1 - e^{-K}).
$$
You can now get the conclusion sending $K\to +\infty$ in the relation
$$
\limsup_{\epsilon\to 0+} \left|
\int_0^\infty f(\epsilon y) e^{-y}\, dy - f(0)
\right|
\leq (|f(0)| + \|f\|_{L^1}) e^{-K}.
$$
