Confusion about the Evaluation of an Integral. I have a confusion about the following question:

Question. $\rho:\mathbb{R} \to \mathbb{R}$ be a continuous function such that $\rho(x) \ge 0 ~\forall ~x \in \mathbb{R}$, and $\rho(x)=0$ if $|x|\ge 1$ and $$\int_{-\infty}^{\infty}\rho(t)dt=1$$ 
  Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function. Evaluate $$\lim_{\epsilon \to 0}\frac{1}{\epsilon}\int_{-\infty}^{\infty}\rho(\frac{x}{\epsilon})~f(x) dx$$

My Solution. The last integral in the question reduces to :
$$
\lim_{\epsilon \to 0}\frac{1}{\epsilon}\int_{-1}^{1}\rho(\frac{x}{\epsilon})~f(x)dx
$$
Now we substitute $x/\epsilon=t$ in the last integral so that we have:
$\lim_{\epsilon \to 0}\frac{1}{\epsilon}\int_{-1}^{1}\rho(\frac{x}{\epsilon})~f(x)dx  = \lim_{\epsilon \to 0}\int_{-1/ \epsilon}^{1/ \epsilon}\rho(t)~f(\epsilon t)dt
 $
Is this right?...First, I cannot understand where the continuity of $\rho$ has been used. Is it even necessary?
Second, Is the interchanging of limit and integral permissible here? or does it require something more for this interchanging?
I'm not sure how to answer these question here...Please help me to understand what is going on here....!! Thank you so much.
 A: First point to be noted: integral with respect to $t$ is from $-\frac 1 \epsilon$ to $-\frac 1 \epsilon$ but since $\rho $ is supported by $[-1,1]$ this does not cause any problem Secondly you have to justify interchnage of limit and integral. Consider $\int_1^{1} \rho (t)f(\epsilon t)\, dt -f(0)$ which you can write as  $\int_1^{1} \rho (t)[f(\epsilon t)-f(0)]\, dt$. Given $\eta >0$ we can find $r$ such that $|f(\epsilon t)-f(0)|<\eta $ if $\epsilon <r$ (because $|\epsilon t | \leq \epsilon)$ Now can you complete the proof?
A: In order to legitimize the interchange of limit and integral, consider the difference
$$\left|\int_{-1/\epsilon}^{1/\epsilon}\rho(t)~f(\epsilon t)dt-f(0)\right|=\left|\int_{-1}^{1}\rho(t)~(f(\epsilon t)-f(0)) dt\right|\leq \int_{-1}^{1}\rho(t)~\left|f(\epsilon t)-f(0)\right| dt$$
where $\epsilon\leq 1$.
Now, by the continuity of $f$ at $0$, for all $n>0$ there is $r>0$ such that $|f(x)-f(0)|<1/n$ for all $|x|<r$. Hence, for $|\epsilon|<r$, then $|\epsilon t|<r$ and
$$\left|\int_{-1}^{1}\rho(t)~f(\epsilon t)dt-f(0)\right|\leq \int_{-1}^{1}\rho(t)(1/n) dt=\frac{1}{n}$$
which goes to $0$ as $n\to+\infty$.
