Evaluate $\lim_{x\to 0}\frac{\sin(\ln(x+1))-\ln(\sin(x)+1)}{\sin^4\frac{x}{2}}$ Compute the following limit: $$L=\lim_{x\to 0}\frac{\sin(\ln(x+1))-\ln(\sin(x)+1)}{\sin^4\left(\frac{x}{2}\right)}$$
 A: 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 $\frac{1/12}{1/16} = \frac{4}{3}$. 
A: Well, it helps to notice that the function doesn't exist for $x\le-1$.
Also, you may also notice that it doesn't exist for $\sin(x)=-1$, which results in $x\ne(2n-1)\pi, n=0,\pm1,\pm2,\pm3,\dots$
The best way I see tackling this is with L'Hospital's rule:
$$L=\lim_{x\to 0}\frac{\sin(\ln(x+1))-\ln(\sin(x)+1)}{\sin^4(\frac{x}{2})}=\lim_{x\to0}\frac{\frac{\cos(\ln(x+1))}{x+1}-\frac{\cos(x)}{\sin(x)+1}}{\cos(\frac{x}2)\sin^3(\frac{x}2)}$$
$$\lim_{x\to0}\frac{\frac{\sin(\ln(x+1))+\cos(\ln(x+1))}{(x+1)^2}+\frac{\sin^2(x)+\sin(x)-\cos^2(x)}{(\sin(x)+1)^2}}{-\frac38\cos(\frac{x}2)\sin(\frac{x}2)\sin^2(\frac{x}2)}$$
It appears obvious that this is only going to get messier... and we pretty much have to repeat this until the limit is evaluable.
At this point, I'd just say your best bet is to look at this graphically or numerically, which produces:
$$L\approx1.333$$
A: $$\begin{align}\sin{\log{(1+x)}} &= \sin{\left (x - \frac12 x^2 + \frac13 x^3 - \cdots \right )} \\ &= \left (x - \frac12 x^2 + \frac13 x^3 - \cdots \right ) - \frac16 \left (x - \frac12 x^2 + \frac13 x^3 - \cdots \right )^3 + \cdots \\ &= x - \frac12 x^2 + \frac16 x^3 + \left ( \frac32 \cdot \frac16 - \frac14 \right ) \frac14 x^4 \cdots\end{align}$$
$$\begin{align}\log{(1+\sin{x})} &= \log{\left (1 + x - \frac16 x^3 +  \cdots \right )} \\ &= \left ( x - \frac16 x^3 +  \cdots \right ) - \frac12 \left (x - \frac16 x^3 +  \cdots \right )^2 + \cdots \\ &= x - \frac12 x^2 + \frac16 x^3 + \left (\frac16-\frac14 \right ) x^4 \cdots\end{align}$$
Thus the limit is equal to
$$\frac{1/12}{1/2^4} = \frac43 $$
A: Hint:
$\sin^4(\frac{x}{2})$~$（\frac{x}{2}）^4$
$\sin(\ln(x+1)) = \ln(x+1)-\frac{\ln^3(x+1)}{6}+o(x^4)$
