Find the limit $\lim_{x\to 0}\frac{\sqrt[5]{1+\sin x}-1}{\ln(1+\tan x)}$

Question

Find the limit $\lim_{x\to 0}\frac{\sqrt[5]{1+\sin x}-1}{\ln(1+\tan x)}$.

I tried subtracting $1$ for using $\lim$ of $e$ but I realized that this is not a problem that can be solved in that way.

Using only algebra, famous limits without L'Hôpital.

Answer 1

Hint:

$$\lim_{x\to0}\dfrac{\sqrt[n]{1+\sin x}-1}{\ln(1+\tan x)}=\lim_{x\to0}\dfrac{\sqrt[n]{1+\sin x}-1}{\sin x}\cdot\lim_{x\to0}\dfrac{\sin x}{\tan x}\cdot\dfrac1{\lim_{x\to0}\dfrac{\ln(1+\tan x)}{\tan x}}$$

For the first limit, set $$\sqrt[n]{1+\sin x}-1=y\implies\sin x=(1+y)^n-1$$

$$\lim_{x\to0}\dfrac{\sqrt[n]{1+\sin x}-1}{\sin x}=\lim_{y\to0}\dfrac y{(1+y)^n-1}=\lim_{y\to0}\dfrac y{ny+O(y^2)}=\cdots=\dfrac1n$$