How to show that a limit exisits, and evaluate the limit of trig functions as $x$ tends to point For each of the following functions $f$, determine whether $\lim_{x\to a}f(x)$
exists, and compute the limit if it exists. In each case, justify your answers.
$$f(x)= x^2\cos\frac{1}{x} (\sin x)^4, \text{ where } a=0$$
$$f(x)= \frac{3(\tan(2x))^2}{2x^2}, \text{ where } a=0$$
I'm awful at trig questions, (I don't think I'm allowed to use L'Hôpital's rule).

Part (a):
$$\lim_{x\to 0}x^2\cos(1/x)(\sin x)^4$$ as $\cos(\mathrm{anything})=$ is defined, but here it is oscillating between $-1$ and $1$ so we can apply limit
$$\lim_{x\to 0}x^2\cos(1/x)(\sin x)^4=(0)*n*(0)$$ where $n=$ some oscillating(but finite) number
hence answer is $0$.
Part (b):
this one is quite straight forward you might be knowing that $\dfrac{\tan x}{x}=1$ when $x\to 0$(why?you can ask in comments)
then similarly $\bigg(\dfrac{\tan x}{x}\bigg)^2=1$ when $x\to 0$ 
so just mutliply and divide you will get$$\lim_{x\to 0}\,\,({3\,\,\text x\,\,2})\bigg(\dfrac{\tan 2x}{2x}\bigg)^2=6$$
