Evaluation of the $\lim_{ x\to 0} (x\ln(\sin(x)))$ without using L'Hopitals Rule? I thought many times but I couldn't find a way to find the $\lim_{x \to 0} x \log(\sin(x)) $ other than the L'Hopitals Rule, Is there any other way to solve it without using L'Hopital Rule?
 A: Hint: I think you know the limit 
$$
\lim\limits_{x\to 0}\frac{\sin(x)}{x}
$$
Then, note that for positive and small enough $x$ we have
$$
x\log(\sin(x))=x\log\left(\frac{\sin(x)}{x}\right)+x\log(x)
$$
Can you proceed from here? Note that it is essential the continuity of $\log$  to get your result from the last expression.
A: You will need the limit of $x\log x$ as $x \to 0^{+}$. This is same as limit of $-\dfrac{\log y}{y}$ as $y \to \infty$ ($y = 1/x$). For this you will need some inequalities. Let $1 < t < y$ and let $a > 0$. Then we know that $$\begin{aligned}&1 > 1 - a \Rightarrow t = t^{1} > t^{1 - a}\\
&\Rightarrow \frac{1}{t} < \frac{1}{t^{1 - a}}\\
&\Rightarrow 0 \int_{1}^{y}\frac{1}{t}\,dt < \int_{1}^{y}\frac{1}{t^{1 - a}}\,dt\\
&\Rightarrow 0 < \log y < \dfrac{y^{a} - 1}{a} < \dfrac{y^{a}}{a}\end{aligned}$$ for $ a > 0, y > 1$. Putting $a = 1/2$ we can easily see that $$0 < \frac{\log y}{y} < \frac{2}{\sqrt{y}}$$ for $y > 1$. Letting $y \to \infty$ and using squeeze theorem we get $$\lim_{y \to \infty}\frac{\log y}{y} = 0$$
Now as in Brandon's answer $$\lim_{x \to 0^{+}}x\log(\sin x) = \lim_{x \to 0^{+}}x\log\left(\frac{\sin x}{x}\right) + x\log x = 0\log 1 + 0 = 0$$
A: The limit does not exist because $\sin(x)$ is negative as you approach zero from the left and you cannot evaluate $\log(x)$ for $ x \leq 0.$
You have to evaluate $$\lim_{x\to 0^+} x\log(\sin(x)) .$$
The difference is small but important.
One thing you could consider is letting $\sin(x) \approx x$. Brandon's post provides a justification for this simplification.
