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

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

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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