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

Evaluate

\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

 

 

 

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