Advertisements
PROBLEM:
Evaluate \displaystyle \lim\limits_{x\to \pi }\left(\frac{\sin^2\left(x\right)}{1+\cos\left(x\right)}\right).
Advertisements
Direct substitution of x=\pi gives the indeterminate form \frac{0}{0}. Therefore, we should apply trigonometric identities.
We know the Pythagorean identity, \sin^2\left(x\right)=1-\cos^2\left(x\right). Therefore, we have
\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*}
Advertisements
Advertisements