Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 18

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PROBLEM:

Evaluate lim⁡x→π(sin⁡2(x)1+cos⁡(x)).\displaystyle \lim\limits_{x\to \pi }\left(\frac{\sin^2\left(x\right)}{1+\cos\left(x\right)}\right).

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SOLUTION:

Direct substitution of x=πx=\pi gives the indeterminate form 00\frac{0}{0}. Therefore, we should apply trigonometric identities.

We know the Pythagorean identity, sin⁡2(x)=1−cos⁡2(x)\sin^2\left(x\right)=1-\cos^2\left(x\right). Therefore, we have

lim⁡x→π(sin⁡2(x)1+cos⁡(x))=lim⁡x→π(1−cos⁡2(x)1+cos⁡(x))=lim⁡x→π((1+cos⁡(x))(1−cos⁡(x))1+cos⁡(x))=lim⁡x→π(1−cos⁡(x))=(1−cos⁡(π))=(1−(−1))=2  (Answer)\begin{align*} \displaystyle \lim\limits_{x\to \pi }\left(\displaystyle \frac{\sin^2\left(x\right)}{1+\cos\left(x\right)}\right) & = \displaystyle \lim\limits_{x\to \pi }\left(\displaystyle \frac{1-\cos^2\left(x\right)}{1+\cos\left(x\right)}\right) \\ \\ & = \displaystyle \lim\limits_{x\to \pi }\left(\displaystyle \frac{\left(1+\cos\left(x\right)\right)\left(1-\cos\left(x\right)\right)}{1+\cos\left(x\right)}\right)\\ \\ & =\lim\limits_{x\to \pi }\left(1-\cos\left(x\right)\right)\\ \\ & =\left(1-\cos\left(\pi \right)\right)\\ \\ & =\left(1-\left(-1\right)\right)\\ \\ & =2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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