# Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 18|

#### $\lim\limits_{x\to \pi }\left(\frac{sin^2\left(x\right)}{1+cos\left(x\right)}\right)$

SOLUTION:

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

We know the pythagorean identity, $sin^2\left(x\right)=1-cos^2\left(x\right)$. So,

$\lim\limits_{x\to \pi }\left(\frac{sin^2\left(x\right)}{1+cos\left(x\right)}\right)=\lim\limits_{x\to \pi }\left(\frac{1-cos^2\left(x\right)}{1+cos\left(x\right)}\right)$

$=\lim\limits_{x\to \pi }\left(\frac{\left(1+cos\left(x\right)\right)\left(1-cos\left(x\right)\right)}{1+cos\left(x\right)}\right)$

$=\lim\limits_{x\to \pi }\left(1-cos\left(x\right)\right)$

$=\left(1-cos\left(\pi \right)\right)$

$=\left(1-\left(-1\right)\right)$

$=2$