I've tried to solve this exercise for a while now but couldn't really find a good way:
Let $\varphi \in C^{\infty}(\mathbb R; [0,1])$ with the following properties:
(i) $\varphi(x) = 0$ for $x\leq -1$; $\quad$ (ii) $\varphi(x) = 1$ for $x\geq 1$; $\quad$ (iii) $\varphi'\geq0$
Given $\varepsilon > 0$, define $\varphi_\varepsilon:\mathbb R \to \mathbb R$; $\quad$ $\varphi_\varepsilon(x):= \varphi(\frac{x}{\varepsilon})$.
Show that, for $f\in C^{\infty}_c(\mathbb R)$ (f has compact support): $$\lim_{\varepsilon \searrow 0} \int_\mathbb R \varphi_\epsilon^2 f'd\lambda^1 = -f(0)$$
I tried using DCT but I am not sure how to deal with the limit even if it is inside the integral. Also I wonder how that $-f(0)$ should possibly come to existence. Any solution/hints appreciated!
$\varphi_\epsilon$ is equal to $0$ on $(-\infty,-\epsilon]$ and is equal to $1$ on $[\epsilon,\infty)$. Then $$ \int_\mathbb R \varphi_\epsilon^2\,f'd\lambda^1=\int_{-\epsilon}^\epsilon\varphi_\epsilon^2\,f'd\lambda^1+\int_\epsilon^\infty f'd\lambda^1=\int_{-\epsilon}^\epsilon\varphi_\epsilon^2\,f'd\lambda^1-f(\epsilon). $$ The integral is bounded by $2(\sup_{x\in\Bbb R}|f'(x)|)\epsilon$, while $\lim_{\epsilon\to0}f(\epsilon)=f(0)$.