How do I solve : $\lim \limits_{x\to0}\frac{6\sin x}{x-3\tan x}$

Question

What is the following limit ? How do I solve it ? $$\lim \limits_{x\to0}\frac{6\sin x}{x-3\tan x}$$

Answer 1

HINT

We don't need l'Hopital indeed we have

$$\lim \limits_{x\to0}\frac{6\sin x}{x-3\tan x}=\lim \limits_{x\to0}\frac{6\frac{\sin x}x}{1-3\frac{\tan x}x}$$

then refer to standard limits.

Answer 2

$$\begin{align*} \lim \limits_{x\to0}\frac{6\sin x}{x-3\tan x} &= \lim \limits_{x\to0} \frac{6\cos(x)}{1 - 3\sec^2(x)} \\ &=\frac{6}{1-3} \\ &= -3\end{align*}$$

Answer 3

$$ \frac{6\sin x}{x-3\tan x} = \frac{6\sin x\cos x}{x(\cos x-3\frac{\sin x}{x})} = 6\cos x\left(\frac{\sin x}{x}\right)\frac{1}{\cos x-3\left(\frac{\sin x}{x}\right)} $$

hence

$$ \lim_{x\to 0}\frac{6\sin x}{x-3\tan x} = \lim_{x\to 0}6\cos x\left(\frac{\sin x}{x}\right)\frac{1}{\cos x-3\left(\frac{\sin x}{x}\right)} = \frac{6}{1-3} = -3 $$

NOTE

We used the fact

$$ \lim_{x\to 0}\frac{\sin x}{x} = 1 $$