defined sequence function :
$$f_{n}(x)=\begin{cases}-n^{2},-\frac{1}{n}<x<0\\n^{2},0<x<\frac{1}{n}\\0,\operatorname{otherwise}\end{cases}$$
Find
$a)$ $\displaystyle\lim_{n\to +\infty}f_{n}(x)$
$b)$ for $\phi \in D$ find the following limit :
$$\displaystyle\lim_{n\to +\infty}\int_{\mathbb{R}}f_{n}(x)\phi (x)dx=?$$
My try for first question :
Fix $x$ , then for all $|x|≥\frac{1}{n}$ then $\lim_{n\to +\infty}f_{n}(x)=0$
Is my solution correct ?
I have a problem always to find this limits of this type definition
And what about second Limit ?
Your calculation of the first limit is correct.
Assuming that you by $\phi \in D$ mean that $\phi$ is differentiable, we can compute the other limit as; $$\lim_{n\rightarrow \infty} \int_{-\infty}^\infty f_n(x) \phi(x) dx = \lim_{n\rightarrow \infty}n^2 (\int_0^{\frac1n}\phi(x)dx-\int_{-\frac{1}{n}}^0\phi(x)dx) $$ Now since $\phi$ is continuous the function $\Phi(y):=\int_0^y \phi(x)dx$ is differentiable with $\Phi'(y)=\phi(y)$ (by the fundamental theorem of calculus). This means that the limit above can be written as $$\lim_{n\rightarrow \infty}n^2 (\int_0^{\frac1n}\phi(x)dx-\int_{-\frac{1}{n}}^0\phi(x)dx) = \lim_{n\rightarrow \infty} \frac{\Phi(\frac{1}{n})+\Phi(-\frac{1}{n})}{\frac{1}{n^2}}$$ Noting that $\Phi(0)=0$ we can use L'Hospitals rule twice to compute $$\lim_{h \rightarrow 0} \frac{\Phi(h)+\Phi(-h)}{h^2}=\lim_{h\rightarrow 0} \frac{\phi(h)-\phi(-h)}{2h} = \lim_{h\rightarrow 0} \frac{\phi'(h)+\phi'(-h)}{2}=\phi'(0) $$