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).

Since the trigonometric functions are in radians, we can say that if $x>0$ then $u < \tan u$, and if $u$ is close enough to $0$, then $\tan u < 1.1u$. Thus with $u=2x$, if $x>0$, we have
$$
1 < \frac{\tan(2x)}{2x} < 1.1.
$$
Since $u \mapsto \dfrac{\tan u} u$ is an even function, this must also hold when $u$ is negative and close enough to $0$.
Then show that what was done with $1.1$ can be done with any other number bigger than $1$ (but how close is "close enough" above depends on how close that number is to $1$).
With all this we can show that $\displaystyle\lim_{x\to0} \frac{\tan(2x)}{2x} = 1$.
But if you're terrible at trig questions, maybe hitting the trigonometry books is what you need to be able to understand this.
