Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 17

PROBLEM:

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SOLUTION:

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*}