Find $\lim_{k \to \infty} f_k$ where the limit is in distribution sense and
- $f_k(x)=k(1-e^{x/k})\delta_{1/k}$
- $f_k(x)=\sin(2 \pi x)\chi_{(k,k+1)}(x)$
When $f_k(x)=k(1-e^{x/k})\delta_{1/k}$
$$\lim_{k \to \infty} f_k=\lim_{k \to \infty} \int_R k(1-e^{x/k})\delta_{1/k}\psi(x)\,dx =\lim_{k \to \infty} k\delta_{1/k} \int_R (1-e^{x/k})\psi(x)\,dx$$
Let $u=x/k \iff uk=x$ and $du=\frac{1}{k}dx$.
$$=\lim_{k \to \infty} k\delta_{1/k} \int_R (1-e^{u})\psi(uk)\frac{du}{k} = \int_R (1-e^{u}) \lim_{k \to \infty} \delta_{1/k} \psi(uk)\,du = \int_R (1-e^{u}) \lim_{k \to \infty} \delta_{1/k} \psi(\infty)\,du = \int_R (1-e^{u}) \lim_{k \to \infty} \delta_{1/k} \times 0 \,du=0$$
and when $f_k(x)=\sin(2 \pi x)\chi_{(k,k+1)}(x)$
$$\lim_{k \to \infty} f_k=\lim_{k \to \infty} \int_R \sin(2 \pi x)\chi_{(k,k+1)}(x)\psi(x)\,dx =\lim_{k \to \infty} \int_k^{k+1} \sin(2 \pi x)\psi(x\,dx$$
Let $u=2 \pi x \iff \frac{u}{2 \pi}=x$ and $du=2 \pi \,dx$
$$=\lim_{k \to \infty} \int \sin(u)\psi(\frac{u}{2 \pi}) \frac{du}{2 \pi}$$
As $\psi(\frac{u}{2 \pi}=\psi(0)+\psi'(\zeta_u)\frac{u}{2 \pi}$, for some $\zeta_u$, we have
$$=\lim_{k \to \infty} \int \sin(u)(\psi(0)+\psi'(\zeta_u)\frac{u}{2 \pi}) \frac{du}{2 \pi}$$
Where do I go from here?
The first computation is correct, but very inefficient. No integration or changes of variable are needed. The distribution $k(1-e^{x/k})\delta_{1/k}$ means: we multiply a test function by $k(1-e^{x/k})$ and then plug in $x=1/k$. This is exactly the same as plugging in $x=1/k$ and multiplying by $k(1-e^{1/k^2})$. The result will not exceed $k(1-e^{1/k^2})\max|\psi|$, and $k(1-e^{1/k^2})\to 0$.
Lesson learned so far: to show something converges to zero, we don't always need to exactly evaluate it; an estimate is enough.
Apply this idea to the second example, using $|\sin |\le 1$.