Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 16|

Evaluate

\lim\limits_{x\to 0}\left(\frac{1-cos^2\left(x\right)}{1+cos\left(x\right)}\right)

SOLUTION:

This problem can be solved using a direct substitution of x=0. That is 

\lim\limits_{x\to 0}\left(\frac{1-cos^2\left(x\right)}{1+cos\left(x\right)}\right)=\frac{1-cos^2\left(0\right)}{1+cos\left(0\right)}

=\frac{1-1}{1+1}

=0

 

 

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