<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] It's easiest to use Taylor expansions about $x = 0$. First the denominator to leading order, so we can tell how many terms are needed:
$$
\sin\left(\frac{x}{2}\right)^4 = \frac{x^4}{16} + O(x^6)
$$
So we need expansions to this order also for the numerator:
$$
\sin(\ln(x+1)) = x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^5}{12}+O\left(x^6\right)
$$
and
$$
\ln(sin(x) + 1) = x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^4}{12}+\frac{x^5}{24}+O\left(x^6\right)
$$
You see that the two terms in the numerator agree through order $x^3$. Hence these terms all cancel. Then the leading term in the numerator is  $\frac{x^4}{12}$. Therefore, the limit equals $\frac{1/12}{1/16} = \frac{4}{3}$. 
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