I had trouble evaluating this limit and found this solution from another post of Stack Exchange. The point of this solution was trying to find the limit without using Taylor's series or L'Hospital's rule.
I just need help understanding certain steps:
How did he go from step 3 to step 4.
How did he go from step 4 to step 5.
His solution:
If we have:
$$4L=\lim_{x\to 0} \frac{\frac12\tan 2x-x}{x^3} \qquad L=\lim_{x\to 0} \frac{\tan x-x}{x^3}$$
substracting we would get
$$4L-L=3L=\lim_{x\to 0} \frac{\frac12\tan 2x-\tan x}{x^3}$$
Now, use that $\tan(2x)=\frac{2\tan x}{1-\tan^2 x}$ to go on:
\begin{equation*} \begin{split} 3L&=\lim_{x\to 0} \frac{\frac{\tan x}{1-\tan^2x}-\tan x}{x^3}=\lim_{x\to 0} \frac{\tan x}{x} \cdot \frac{\frac{1}{1-\tan^2 x}-1}{x^2}=\lim_{x\to 0} \frac{\tan x}{x}\cdot \frac{\tan^2 x}{x^2(1-\tan^2 x)}=\\=&\lim_{x\to 0} \frac{\tan x}{x}\cdot \frac{\tan^2 x}{x^2}\cdot\frac{1}{1-\tan^2 x}=\lim_{x\to 0} \frac{\tan^3 x}{x^3} \cdot \lim_{x\to 0} \frac{1}{1-\tan x^2}\end{split} \end{equation*}
Now, as $\lim_{x\to 0} \frac{1}{1-\tan^2 x}=1$, (because $\tan x\to 0$ when $x\to 0$) and $$\lim_{x\to 0} \frac{\tan^3 x}{x^3}=\left(\lim_{x\to 0} \frac{\tan x}{x}\right)^3=1^3=1$$ we get that $3L=1$.