How to work with trig functions when dealing with limits tending to a point, without using L'Hôpital's rule 
Past Paper Question:

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.
a) $f(x)= \dfrac{2\tan^2(x)}{x^2+2x^3}$ $\text{where a = 0}$
b)$f(x)= \dfrac{\sin(\left| x \right|)}{x^2+\sin(\left| x \right|)}$ $\text{where a = 0}$

Attempt:
I can't use L'Hôpital's rule, but I know from my graphical calculator that the limits of a) = $6$ and b) = $1$. 

Not sure how to show this, from past experience I know you have to separate the $\tan(x)$ function into $\cos(x)$ and $\sin(x)$. But apart from that I have no idea. 
 A: Use equivalents:
$\tan x\sim_0 x$, $x^2+2x^3\sim_0 x^2$, hence
$$\frac{2\tan^2 x}{x^2+2x^3}\sim_0\frac{2x^2}{x^2}=2\quad\text{(not }6)$$
$\sin\lvert x\rvert\sim_0\lvert x\rvert$, one checks readily that $x^2+\sin\mkern1mu\lvert x\rvert\sim_0\lvert x\rvert$, so
$$\dfrac{\sin(\left| x \right|)}{x^2+\sin(\left| x \right|)}\sim_0\frac{\lvert x \rvert}{\lvert x \rvert}=1.$$
A: You don't need series. 
Hints:
a). $\dfrac{2\tan^2(x)}{x^2+2x^3}=2\left ( \frac{\tan^{2}x}{x^{2}} \right )\left ( \frac{1}{1+2x} \right )$
b). $\dfrac{\sin(\left| x \right|)}{x^2+\sin(\left| x \right|)}=\frac{\frac{\sin(\left| x \right|)}{\vert x\vert}}{\frac{x^2+\sin(\left| x \right|)}{\vert x\vert}}$
A: You should be able to figure out if the limit exists or not, the problem is when trying to evaluate it
According to wolfram the first limit is actually 2. We can evaluate the first limit using the fundamental trig limit $\lim_{x \to 0} \frac{\tan x}{x} = 1 $. So 
\begin{align}
\lim_{x \to 0} \frac{2\tan^2 x}{x^2 + 2x^3} &= \lim_{x \to 0} (\frac{\tan x}x{})^2 \frac{2}{1 + 2x}\\
&= \lim_{x \to 0} 1 \cdot \frac{2}{1 + 2x}\\
&= 2
\end{align}
As for the second limit, recall the fundamental limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$. An equivalent form of this limit is $\lim_{x \to 0} \frac{\sin |x|}{|x|}$. Since $x^2 = |x||x|$(as the square is positive) we  can factor
\begin{align}
\lim_{x \to 0} \frac{\sin|x|}{x^2 + \sin|x|} &= \lim_{x \to 0} \frac{1}{|x|} \frac{\sin|x|}{|x| + \frac{\sin|x|}{|x|}}\\
&= \lim_{x \to 0} \frac{\frac{\sin|x|}{|x|}}{|x|  + \frac{\sin|x|}{|x|}}\\
&= \lim_{x \to 0} \frac{1}{1 + |x|}\\
&= 1 
\end{align}
