$\lim_{x \to 0^+} x \log\{\sin(x) \}=?$

Question

How can I evaluate the right hand limit of $x \log\{\sin(x) \}$ as $x \to 0$ ; that is: $$\lim_{x \to 0^+} x \log\{\sin(x) \}=?$$

My Approach:
Let $f(x)= x \log\{\sin(x) \}$
then, $f(0.1)=0.1 \log\{\sin(0.1) \}$ ; and we know for small values of $x : \quad \sin(x) \approx x$
so, $f(0.1)=0.1 \log\{0.1 \}=0.1\log\{10^{-1} \}=-0.1$
similarly, $f(0.01)=0.01 \log\{\sin(0.01) \}=0.01 \log\{0.01 \}=0.01\log\{10^{-2} \}=-0.02$
so we conclude as $x \to 0$ , $f(x) \to 0$
Therefore , $\lim_{x \to 0^+} x \log\{\sin(x) \}=0$

But, how can we find it in a systematic process? please help...

Answer 1

\begin{align} \lim_{x \to 0^+} x \log(\sin(x)) &= \lim_{x \to 0^+} \frac{\log(\sin x)}{\frac1x}\\ &= \lim_{x \to 0^+} -\frac{x^2\cos x}{\sin x}\\ &= -\lim_{x \to 0^+ }x \cos x \\&=0 \end{align}

Answer 2

\begin{align*} \lim_{x\rightarrow 0^{+}}x\log(\sin x)&=\lim_{x\rightarrow 0^{+}}\dfrac{\log(\sin x)}{1/x}\\ &=\lim_{x\rightarrow 0^{+}}\dfrac{(1/\sin x)\cos x}{-1/x^{2}}\\ &=\lim_{x\rightarrow 0^{+}}-\dfrac{x}{\sin x}\cdot(x\cos x)\\ &=-1\cdot 0\\ &=0. \end{align*}

Answer 3

Using equivalents for small values of $x$ $$\sin(x) \sim x \implies \log(\sin(x)) \sim \log(x)\implies x\log(\sin(x)) \sim x\log(x)$$

Using Taylor series $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)=x \left(1-\frac{x^2}{6}+O\left(x^4\right) \right)$$ $$\log(\sin(x))=\log(x)+\log\left(1-\frac{x^2}{6}+O\left(x^4\right) \right)=\log(x)-\frac{x^2}{6}+O\left(x^4\right)$$ $$x\log(\sin(x))=x\log(x)-\frac{x^3}{6}+O\left(x^4\right)$$

Answer 4

By standard limits

$$x \log(\sin(x))=\frac{x}{\sin x}\cdot \sin x\cdot \log(\sin(x))\to 1\cdot 0=0$$