For what value of $\alpha$ is the limit $\lim{x\to 0} \frac{\arctan(x) \cdot \log(\sin(x)) - x \cdot \log(x) }{x^\alpha}$ finite and non-zero? My brother asked me this question and I found it very difficult.
For which value of $\alpha$ is the limit 
$$
\lim_{x\to0} \frac{\arctan(x) \cdot \log(\sin(x)) - x \cdot \log(x) }{x^\alpha}
$$ finite and different from zero? The thing that confuses me is that I can't use McLaurin for $\log x$, because it is not defined in 0.
Is the question really hard or am I missing something?
 A: You can expand $\log (\sin(x))=\log(x)-\frac{x^2}{6}$ and $\tan^{-1}(x)=x-\frac{x^3}{3}$ plus higher order terms.
The $x\log x$ terms cancel, but the expression still goes to 
$$-x^{3-\alpha}\left(\frac16 + \frac13 \log(x)\right)$$
This goes to zero for $\alpha < 3$ and to infinity for $\alpha \geq 3$ so your problem has no answer.
A: It is possible to arrive to the same conclusion as that of Mark Fischler,
without using Taylor series. First note that
\begin{equation*}
\log (\sin x)=\log \left( x\frac{\sin x}{x}\right) =\log x+\log \left( \frac{%
\sin x}{x}\right) =\log x+\log \left( 1+\frac{\sin x-x}{x}\right) .
\end{equation*}
Then,
\begin{eqnarray*}
\frac{\arctan x\log (\sin x)-x\log x}{x^{\alpha }} &=&\frac{\arctan x\left(
\log x+\log \left( 1+\frac{\sin x-x}{x}\right) \right) -x\log x}{x^{\alpha }}
\\
&=&\frac{\left( \arctan x-x\right) \log x+\arctan x\log \left( 1+\frac{\sin
x-x}{x}\right) }{x^{\alpha }} \\
&=&\left( \frac{\arctan x-x}{x^{3}}\right) \left( \frac{\log x}{x^{\alpha -3}%
}\right)  \\
&&+\left( x^{3-\alpha }\right) \left( \frac{\arctan x}{x}\right) \left( 
\frac{\sin x-x}{x^{3}}\right) \left( \frac{\log \left( 1+\left( \frac{\sin
x-x}{x^{3}}\right) x^{2}\right) }{\left( \frac{\sin x-x}{x^{3}}\right) x^{2}}%
\right) 
\end{eqnarray*}
Using standard limits 
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6},\ \ \ \ \ \ \
\lim_{x\rightarrow 0}\frac{\arctan x}{x}=1,\ \ \ \ \ \ \ \
\lim_{x\rightarrow 0}\frac{\arctan x-x}{x^{3}}=-\frac{1}{3},\ \ \ \
\lim_{u\rightarrow 0}\frac{\log (1+u)}{u}=1 \\
\lim_{x\rightarrow 0^{+}}\frac{\log x}{x^{\alpha -3}} &=&\left\{ 
\begin{array}{ccc}
0 & if & \alpha <3 \\ 
-\infty  &  & \alpha =3 \\ 
+\infty  &  & \alpha >3%
\end{array}%
\right. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\lim_{x\rightarrow 0^{+}}x^{3-\alpha }=\left\{ 
\begin{array}{ccc}
0 & if & \alpha <3 \\ 
1 &  & \alpha =3 \\ 
+\infty  &  & \alpha >3%
\end{array}%
\right. 
\end{eqnarray*}
it follows that
\begin{equation*}
\lim_{x\rightarrow 0^{+}}\frac{\arctan x\log (\sin x)-x\log x}{x^{\alpha }}%
=\left\{ 
\begin{array}{ccc}
\left( -\frac{1}{3}\right) \left( 0\right) +\left( 0\right) \left( 1\right)
\left( -\frac{1}{6}\right) \left( 1\right) =0 & if & \alpha >3 \\ 
\left( -\frac{1}{3}\right) \left( -\infty \right) +\left( 1\right) \left(
1\right) \left( -\frac{1}{6}\right) \left( 1\right) =+\infty  & if & \alpha
=3 \\ 
\left( -\frac{1}{3}\right) \left( +\infty \right) +\left( +\infty \right)
\left( 1\right) \left( -\frac{1}{6}\right) \left( 1\right) =-\infty  & if & 
\alpha <3.%
\end{array}%
\right. 
\end{equation*}
For any real value of $\alpha ,$ the limit is either infinite or
zero.
