How to solve the limit of $\ln(x \sin x)$ as $x$ approaches $0$? I have a limit:
$$k = \lim_{x\to0+} \ln(x \sin x)$$
How do I find this? Since $\ln(x)$ is continuous I tried:
$$k = \ln( \lim_{x\to0+}  (x \sin x))$$
$$k = \ln(0)$$
Which to my understanding is undefined, but the answer is $-\infty$ somehow. Where did I go wrong? 

You could use Taylor series for approximating the value of $\log (x \sin (x))$ and then get the limit (and how it is approached).
$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$
$$x\sin(x)=x^2-\frac{x^4}{6}+\frac{x^6}{120}+O\left(x^8\right)=x^2\left(1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right) \right)$$
$$\log (x \sin (x))=\log(x^2)+\log\left(1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right) \right)$$
$$\log (x \sin (x))=\log(x^2)-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right)$$
Just for curiosity, make $x=\frac \pi 6$. The exact value is
$$\log \left(\frac{\pi }{12}\right)\approx -1.34018$$ while the approximation would give
$$2\log\left(\frac \pi 6\right)-\frac{\pi ^2}{216}-\frac{\pi ^4}{233280}\approx -1.34017$$
