How do I evaluate this limit using L'Hospitals Rule

Question

Evaluate:$${\lim_{x \to 0}} \frac{\tan^2x-x^2}{x^2\tan^2x}$$

I tried it using L'Hospital's Rule but the process seems to be non-terminating.

Any hints please?

Answer 1

$$\lim_{x\to0}\dfrac{\tan^2x-x^2}{x^2\tan^2x}=\lim_{x\to0}\dfrac{x+\tan x}x\cdot\left(\lim_{x\to0}\dfrac x{\tan x}\right)^2\cdot\lim_{x\to0}\dfrac{\tan x-x}{x^3}$$

First two limits are elementary, right?

Now apply L'Hosiptal on $\lim_{x\to0}\dfrac{\tan x-x}{x^3}$

or use Are all limits solvable without L'Hôpital Rule or Series Expansion

Answer 2

Another way:

$$ \frac{1}{x^2} - \frac{1}{\tan^2 x} = \frac{1}{x^2} + 1 - \frac{1}{\sin^2 x} $$

Instead of L'Hospital's rule, you can expand $\sin$ as $x \to 0: \sin x \sim x - \frac{x^3}{3!}$. When you plug in this expansion in the third fraction you get $$ \frac{1}{\bigg(x - \frac{x^3}{3!} \bigg)^2} = \frac{1}{x^2 \bigg( 1 - \frac{x^2}{6!} \bigg)^2} $$ so the whole expression becomes $$ \frac{1}{x^2} \bigg(1 - \frac{1}{(1- \frac{x^2}{6})^2}\bigg) + 1 $$
Now expand the brackets, cancel out $x^2$ and get $- \frac{1}{3} + 1 = \frac{2}{3}$