<s>[INST] Evaluate $\lim_{x\to 0}\frac{\sin(\ln(x+1))-\ln(\sin(x)+1)}{\sin^4\frac{x}{2}}$ Compute the following limit: $$L=\lim_{x\to 0}\frac{\sin(\ln(x+1))-\ln(\sin(x)+1)}{\sin^4\left(\frac{x}{2}\right)}$$
 [/INST] $$\begin{align}\sin{\log{(1+x)}} &= \sin{\left (x - \frac12 x^2 + \frac13 x^3 - \cdots \right )} \\ &= \left (x - \frac12 x^2 + \frac13 x^3 - \cdots \right ) - \frac16 \left (x - \frac12 x^2 + \frac13 x^3 - \cdots \right )^3 + \cdots \\ &= x - \frac12 x^2 + \frac16 x^3 + \left ( \frac32 \cdot \frac16 - \frac14 \right ) \frac14 x^4 \cdots\end{align}$$
$$\begin{align}\log{(1+\sin{x})} &= \log{\left (1 + x - \frac16 x^3 +  \cdots \right )} \\ &= \left ( x - \frac16 x^3 +  \cdots \right ) - \frac12 \left (x - \frac16 x^3 +  \cdots \right )^2 + \cdots \\ &= x - \frac12 x^2 + \frac16 x^3 + \left (\frac16-\frac14 \right ) x^4 \cdots\end{align}$$
Thus the limit is equal to
$$\frac{1/12}{1/2^4} = \frac43 $$
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