Trouble solving this limit without using l'Hôpital's rule

Question

The limit I want to calculate is the following $$ \lim_{x \to 0}{\frac{(e^{\sin(4x)}-1)}{\ln\big(1+\tan(2x)\big)}} $$ I've been stuck on this limit for a while and I don't know how to solve it please help me.

Answer 1

Hint: $$\frac{(e^{\sin 4x}-1)}{\ln(1+\tan 2x)}= \frac{(e^{\sin 4x}-1)}{\sin 4x}\frac{\tan 2x}{\ln(1+ \tan 2x)} \frac{\sin 4x}{4x} \frac{2x}{\tan 2x} \times 2.$$

Answer 2

As $x\to 0$, $$\exp(\sin 4x)-1\sim\sin4x\sim 4x$$ and $$\ln(1+\tan 2x)\sim\tan2x\sim 2x$$ etc.

Answer 3

Hint: Make use the following facts: 1. $\dfrac{e^{\sin(4x)} - 1}{\sin(4x)} \to 1$

2.$\dfrac{\log(1 + \tan(2x))}{\tan(2x)} \to 1$

3. $\cos(2x) \to 1$