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PROBLEM:
Evaluate \displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\sin\left(x\right)\sin\left(2x\right)}{1-\cos\left(x\right)}\right).
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Direct substitution of x=0 gives the indeterminate form \frac{0}{0}. Therefore, we should apply trigonometric identities.
We know that \sin\left(2x\right)=2\sin\left(x\right)\cos\left(x\right), so we can rewrite the original function as
\begin{align*}
\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\sin\left(x\right)\sin\left(2x\right)}{1-\cos\left(x\right)}\right) & =\lim\limits_{x\to 0}\left(\frac{\sin\left(x\right)\cdot 2\left(\sin\left(x\right) \cos\left(x\right)\right)}{1-\cos\left(x\right)}\right)\\
\\
& =\displaystyle 2\cdot \lim\limits_{x\to 0}\left(\frac{\sin^2\left(x\right)\cos\left(x\right)}{1-\cos\left(x\right)}\right)\\
\end{align*}We also know the Pythagorean identity \sin^2\left(x\right)=1-\cos^2\left(x\right). So,
\begin{align*}
\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\sin\left(x\right)\sin\left(2x\right)}{1-\cos\left(x\right)}\right) & =2\cdot \lim\limits_{x\to 0}\left(\frac{\left(1-\cos^2\left(x\right)\right)\cos\left(x\right)}{1-\cos\left(x\right)}\right)\\
\\
& =2\cdot \lim\limits_{x\to 0}\left(\frac{\left(1+\cos\left(x\right)\right)\left(1-\cos\left(x\right)\right)\cos\left(x\right)}{1-\cos\left(x\right)}\right) \\
\\
& =2\cdot \lim\limits_{x\to 0}\left(\left(1+\cos\left(x\right)\right)\cos\left(x\right)\right) \\
\\
& = 2\cdot \left(\left(1+\cos\left(0\right)\right)\cos\left(0\right)\right) \\
\\
& =2\cdot \left(\left(1+1\right)\cdot 1\right) \\
\\
& =4 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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