$\psi,f:\Bbb R\to\Bbb R$ are continuous, $\psi=0$ outside $[0,1]$, $\int\psi=1$. $\lim_{n\to\infty}n\int_0^{100}f(x)\psi(nx)dx=?$ [duplicate]

The QuestionIf $\psi:\mathbb{R} \rightarrow \mathbb{R}$ is a continuous function $\ni$$\int_{0}^{1}\psi(y)dy = 1$, $\psi(y) = 0$ when $y$ doesn't belong to $[0,1]$ and $f: \mathbb{R} \rightarrow \mathbb{R}$ is a twice differentiable function Then, calculate the following limit.

$$\lim_{n\to\infty}n\int_0^{100}f(x)\psi(nx)dx$$

My Attempt:Firstly, I restricted the limits from $0$ to $\frac{1}{n}$ as $\psi(nx)$ is $0$ outside 1. Then I used L'Hôpital's rule, the above limit equals:

$\lim_{n \to \infty}\frac{f(\frac{1}{n})\psi(1)(n^{-2})-f(0)\psi(0)\cdot0}{-\frac{1}{n^{2}}}$ (using the formula of differentiating under integral sign) which yields $f(0)\psi(1)$. But the answer is expected to come only in terms of $f$ or $f^{'}$ on some point.

I tried By parts as well, but that didn't help as well. Kindly guide me here.