Need help understanding certain steps of the solution to a limit.

I had trouble evaluating this limit and found this solution from another post . 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:

1. How did he go from step 3 to step 4.




2. How did he go from step 4 to step 5.

His solution:

Answer

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 $3L=1$.