Integrating unit impulse function Given that, 
$$ \delta(t) = \begin{cases} \infty & \text{if } t = 0 \\ 
0 & \text{if } t \ne 0\\ 
\end{cases}$$
How is it that, 
(A) 
$$
\int_{-\infty}^\infty \delta(t) dt = 1
$$
(B)
$$
\int_{-\infty}^\infty f(t) \delta(t) dt = f(0)
$$
considering $f$ continuos at $t=0$
Thanks in advance

One way to see this is by approximating the Dirac delta function $\delta(t)$ by a sequence of continuous functions. A convenient choice of approximating functions is the set of Gaussian kernels (i.e. normal distributions) with decreasing variance. 
Let,
\begin{align}
f_n(t) &= \frac{1}{\sigma_n \sqrt{2\pi}} e^{-\frac{t^2}{2\sigma_n^2}},
\end{align}
for positive integers $n$. We let $\sigma_n = \frac{1}{n}$, so that the variance approaches zero as $n$ grows arbitrarily large.
You can see that a the sequence $\{f_n\}$ converges pointwise to the Dirac delta function $\delta(t)$. There are some subtle points that I will not bring up here; but, if you plot this sequence, you will at least gain some intuition.
Since the sequence $\{f_n\}$ is a sequence of probability distributions, we have:
\begin{align}
\lim_{n \to \infty} \int_{-\infty}^{\infty}{f_n(t)dt} &= \lim_{n \to \infty} \int_{-\infty}^{\infty} {\frac{1}{\sigma_n \sqrt{2\pi}} e^{-\frac{t^2}{2\sigma_n^2}}} \\
&= \lim_{n \to \infty} 1 \\
&= 1.
\end{align}
Again, I have omitted some details here. To understand this in full mathematical rigor, you should take a course in measure theory.
