Let $\rho$ : $\mathbb{R}$ $\rightarrow$ $\mathbb{R}$ be a continuous function such that Let $\rho$ : $\mathbb{R}$ $\rightarrow$ $\mathbb{R}$ be a continious function such that $\rho(x) \geqslant 0 $ for all $x \in \mathbb{R}$, $\rho =0 $ if 
$\mid x \mid \geqslant 1$ and
$$\int_{-\infty}^{\infty} \rho(t) dt = 1.$$
Let $f(x): \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function.
Evaluate $$ \lim_{\epsilon \to 0} \frac{1}{\epsilon} \int_{-\infty}^{\infty}
\rho (\frac{x}{\epsilon})f(x)dx .$$ 
I have constructed such a function $\rho(x)$ which satisfies the the condition given above, but I am unable to compute the limit for my constructed function. Kindly help me. My constructed function $\rho(x)$ is 
$$\rho(x) =\left\{\begin{array}{ll}
        x+1   & \text{for } x \in [-1,0],\\
        -x+1   & \text{for } x \in [0,1],\\
        0 & \text{otherwise.} 
        \end{array}\right. 
$$
 A: $\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\int_{-\infty}^{\infty}\rho(\frac{x}{\varepsilon})f(x))=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\int_{-\varepsilon}^{\varepsilon}\rho(\frac{x}{\varepsilon})f(x))$
Denote the integral inside the limit by $I_\varepsilon$
$m_\varepsilon\int_{-\varepsilon}^\varepsilon\rho(\frac{x}{\varepsilon})dx\leq I_\varepsilon \leq M_\varepsilon\int_{-\varepsilon}^\varepsilon\rho(\frac{x}{\varepsilon})dx$
where $m_\varepsilon$ and $M_\varepsilon$ are the bounds on $f$ on the interval $[-\varepsilon;+\varepsilon]$
The integral we have is obviously $\varepsilon$ and thus 
$m_\varepsilon\varepsilon\leq I_\varepsilon \leq M_\varepsilon\varepsilon$
Divide both sides by $\varepsilon$, note that $m_\varepsilon$ and $M_\varepsilon$  go to $f(0)$ by continuity of $f$, and we are done. The limit is $f(0)$.
A: Hint: The expression equals $\int_{-1}^1 \rho (y)f(\epsilon y)\, dy.$
A: Another hint: Use that $f$ is continuous and look closely into:
$$ \int_{\Bbb R} \frac{1}{\epsilon} \rho\left( \frac{x}{\epsilon} \right) \left( f(x) -f (0) \right) \; dx$$
