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!
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 $$ \limsup_{\epsilon\to 0+} \left| \int_0^\infty f(\epsilon y) e^{-y}\, dy - f(0) \right| \leq (|f(0)| + \|f\|_{L^1}) e^{-K}. $$