how to prove that: $\lim_{n\to \infty} \int_{-\infty}^{+\infty}\,f(t)\cdotδ_n(t)\,dt=f(0)$ using limit of a sequence? The limit of a sequence declares that: $\lim_{x\to \infty}a_n=L$ if and only if for some$\,n>N_o, |a_n-L|<ε$ where $ε>0, N_0>0$
I want to use that to prove:
$\lim_{n\to \infty} \int_{-\infty}^{+\infty}\,f(t)\cdotδ_n(t)\,dt=f(0)$, $f(t)$ is continuous at $t=0$.
$${\rm Definition:} \qquadδ_n(t)=\Big\{^{n,\,0<t<\frac1n}_{0,\,elsewhere}$$ 
$$\Big|f(t)-f(0)\Big|<ε \Leftarrow \Rightarrow$$
$$\Big|\int_{0}^{\frac1n}\,nf(t)dt\,-\,f(0)\Big|<ε \Leftarrow \Rightarrow$$
$$\Big|\int_{0}^{\frac1n}\,n\Big(f(t)\,-\,f(0)\Big)\,dt\Big|<ε \Leftarrow \Rightarrow$$
Now i am stuck. I do not see what are the logic steps to solve it. 
 A: We want to show that for all $\epsilon>0$ there is an $N_0\in\mathbb{N}$ such that for all $n>N_0$ we have,
$$\left|\int\limits_{-\infty}^\infty f(t)\delta_n(t)dt - f(0)\right|<\epsilon.$$
Let $\epsilon>0$. Then,
$$\left|\int\limits_{-\infty}^\infty f(t)\delta_n(t)dt - f(0)\right| = \left|\int\limits_{0}^{\frac{1}{n}} nf(t)dt - f(0)\right|=\left|\int\limits_{0}^{\frac{1}{n}} n(f(t)- f(0))dt \right|\leq \int\limits_{0}^{\frac{1}{n}} \left|n(f(t)- f(0))\right|dt \leq \frac{1}{n}\sup\limits_{t\in(0,\frac{1}{n})}\left(n |f(t)-f(0)|\right)=\sup\limits_{t\in(0,\frac{1}{n})}|f(t)-f(0)|$$
Now we need some assumption on $f$ to finish (continuity?).
Edit: with continuity we can bound $|f(t)-f(0)|$ for $t\in(0,\frac{1}{n})$. Since $f$ is continuous, we have for our choice of $\epsilon$ that there exists $\delta>0$ such that,
$$|t-0|<\delta\implies |f(t)-f(0)|<\epsilon$$
Now, we choose $N_0$ such that for all $n>N_0$ we have,
$$\frac{1}{n}<\delta\implies\sup\limits_{t\in(0,\frac{1}{n})}|f(t)-f(0)|\leq \sup\limits_{t\in(0,\delta)}|f(t)-f(0)|$$
A: You have almost got the proof. All you have to do is split the integral into two separate terms. In your last line $\int_0^{n} nf(t)dt =\int _{-\infty} ^{\infty} f(t) \delta_n (t) dt$ and $\int_0^{n} nf(0))dt =f(0)$. That finishes the proof, right?
