How to solve the limit of $\ln(x \sin x)$ as $x$ approaches $0$?

Question

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?

Answer 1

Use the fact that $\log u\rightarrow-\infty$ as $u\rightarrow 0^{+}$. Then given $M<0$, we have some $\delta>0$ such that for all $u\in(0,\delta)$, then $\log u<M$.

Now use the fact that $x\sin x\rightarrow 0$ as $x\rightarrow 0^{+}$, then we have some $\delta'\in(0,\pi/2)$ such that $x\sin x=|x\sin x|<\delta$ for all $x\in(0,\delta')$, then $\log(x\sin x)<M$ for all such $x$, this shows that $\lim_{x\rightarrow 0^{+}}\log(x\sin x)=-\infty$.

Answer 2

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 $$2\log\left(\frac \pi 6\right)-\frac{\pi ^2}{216}-\frac{\pi ^4}{233280}\approx -1.34017$$

Answer 3

Let $y=x \sin x$ an note that for $x>0$ we have that $x \sin x>0\implies y=x \sin x \to0^+$, thus

$$\lim_{x\to0+} \ln(x \sin x)=\lim_{y\to0+} \ln y=-\infty$$