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 thought that it would be instructive to present a way forward that circumvents the use of L'Hospital's Rule, asymptotic analysis, or other derivative-based methodologies.  To that end, herein, we use some basic inequalities and the squeeze theorem.  
In THIS ANSWER for $x>-1$, I showed
$$\frac{x}{x+1}\le\log (1+x)\le x \tag 2$$
using only Bernouli's Inequality and the limit definition of the exponential function.
And here, the inequality 
$$|x\cos x|\le |\sin x|\le |x| \tag 1$$
was established by appealing to geometry only.
Using $(1)$ and $(2)$, we have for $x>0$
$$\frac{(1+x\cos x)^{1/5}-1}{\tan x}\le\frac{(1+\sin x)^{1/5}-1}{\log (1+\tan x)}\le\frac{(1+x)^{1/5}-1}{\frac{\tan x}{1+\tan x}} \tag 3$$
Note that the right-hand side 0f $(3)$ can be written
$$\begin{align}
\frac{(1+x)^{1/5}-1}{\frac{\tan x}{1+\tan x}}& =(1+\tan x)\left(\frac{x\cos x}{\sin x}\right)\\\\\
&\times \left(\frac{1}{1+(1+x)^{1/5}+(1+x)^{2/5}+(1+x)^{3/5}+(1+x)^{4/5}}\right)\\\\
&\to \frac 15\,\,\text{as}\,\,x\to 0 
\end{align}$$
Similarly, the left-hand side 0f $(3)$ can be written
$$\begin{align}
\frac{(1+x\cos x)^{1/5}-1}{\tan x}& =(\cos^2 x)\left(\frac{x}{\sin x}\right)\\\\
&\times \left(\frac{1}{1+(1+x\cos x)^{1/5}+(1+x\cos x)^{2/5}+(1+x\cos x)^{3/5}+(1+x\cos x)^{4/5}}\right)\\\\
&\to \frac 15\,\,\text{as}\,\,x\to 0 
\end{align}$$
Therefore, by the squeeze theorem 
$$\lim_{x\to 0^+}\frac{(1+\sin x)^{1/5}-1}{\log (1+\tan x)}=\frac15$$
A similar development for $x<0$ results in the same limit.  Therefore, we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{(1+\sin x)^{1/5}-1}{\log (1+\tan x)}=\frac15}$$
and we are done without using anything other than standard inequalities!
