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?
 A: Use Taylor's developments at order $1$ and compose them:
\begin{align*}&\left.\begin{array}{l}
\sin x=x+o(x)\\\sqrt[5]{1+u}=1+ \dfrac15u+o(u)
\end{array}\right\}\Rightarrow\enspace\sqrt[5]{1+\sin x}-1= 1+ \frac15x+o(x)-1=\frac15x+o(x)\\[1ex]
&\left.\begin{array}{l}
\tan x=x+o(x)\\\ln(1+u)=u+o(u)
\end{array}\right\}\Rightarrow\enspace\ln{1+\tan x}= x+o(x)
\end{align*}
whence
$$\frac{\sqrt[5]{1+\sin x}-1}{\ln(1+\tan x)}=\frac{\dfrac15x+o(x)}{x+o(x)}=\frac{\dfrac15+o(1)}{1+o(1)}\to \frac15.$$
A: 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!
A: 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}
