Unit Impulse funciton FT How is FT of $ \delta $(t) equal to 1 ? Normal FT gives the result $\infty$. Can someone please explain? I did the normal integration and substituted the limits. 

Is it because  $ \delta $(t) is a unit impulse function so as it's height is large it's width is very small so no matter what the FT will always be equal to 1 ? (I'm just trying to figure out the logic)
If follows directly from the definition of the Dirac delta distribution. It is defined so that

$$\int_\mathbb{R} \delta(x) f(x) dx := f(0)$$

for any test function $f(x)$. In other words, the Dirac distribution is the generator of the linear functional that extracts a single function value.

With this definition the Fourier transform of the Dirac distribution is simply:
$$\int_\mathbb{R} \delta(t) \exp(-2\pi i \omega t) dt=\exp(-2\pi i \omega\cdot 0)=\exp(0)=1$$

It doesn't make much sense to say the Dirac distribution is infinitely high or infinitely narrow. Just use the definition given above and apply it. If you really need to, you can understand the distribution as the limit of a sequence of certain functions, but it's not a function itself. And specifically it doesn't have a graph.