Evaluate the limit without using the L'Hôpital's rule $$\lim_{x\to 0}\frac{\sqrt[5]{1+\sin(x)}-1}{\ln(1+\tan(x))}$$
How to evaluate the limit of this function without using L'Hôpital's rule?

I'm not sure if this can be made formal...
$$\lim_{x\to 0}\frac{\sqrt[5]{1+\sin(x)}-1}{\ln(1+\tan(x))}$$

$$\sqrt[5]{1+x} = 1 + \dfrac x5 - \dfrac{2}{25}x^2 + O(x^3)$$
$$\sqrt[5]{1+\sin(x)}-1 \approx \dfrac 15 \sin x - \dfrac{2}{25}\sin{x^2}$$

$$\ln(1+x) = x - \dfrac 12x^2 + O(x^3)$$
$$\ln(1+ \tan x) \approx \tan x - \dfrac 12 \tan^2 x $$

\begin{align}
   \frac{\sqrt[5]{1+\sin(x)}-1}{\ln(1+\tan(x))} 
   & \approx 
     \frac{\dfrac 15 \sin x-\dfrac{2}{25}\sin{x^2}}{\tan x-\dfrac 12 \tan^2 x} \\
   & \approx 
     \frac{\dfrac 15 \sin x-\dfrac{2}{25}\sin{x^2}}{\tan x-\dfrac 12 \tan^2 x} \\
   & \approx
     \dfrac{\sin x}{\tan x}
     \frac{\dfrac 15 - \dfrac{2}{25}\sin x}{1 - \dfrac 12 \tan x} \\
   & \approx
     \cos x \,
     \frac{\dfrac 15 - \dfrac{2}{25}\sin x}{1 - \dfrac 12 \tan x} \\
   & \to \dfrac 15 \quad \text{as} \quad x \to 0
\end{align}
