Limit of [1 - cos (x)]/x as x --> 0

**khamaar** · June 30th 2011, 04:45 AM

WHAT IS LIMIT OF , [1 - cos (x)]/x such that "x approaches zero"

i think the answer is "limit doesnt exist"..

The book says its 0.

**e^(i*pi)** · June 30th 2011, 04:55 AM

Are you allowed to use L'Hopital's rule?

**Prove It** · June 30th 2011, 04:59 AM

Originally Posted by khamaar

WHAT IS LIMIT OF , [1 - cos (x)]/x such that "x approaches zero"

i think the answer is "limit doesnt exist"..

The book says its 0.

$\displaystyle \begin{align*} \lim_{x \to 0}\frac{1 - \cos{x}}{x} &= \lim_{x \to 0}\frac{(1 - \cos{x})(1 + \cos{x})}{x(1 + \cos{x})} \\ &= \lim_{x \to 0}\frac{1 - \cos^2{x}}{x(1 + \cos{x})} \\ &= \lim_{x \to 0}\frac{\sin^2{x}}{x(1 + \cos{x})} \\ &= \lim_{x \to 0}\frac{\sin{x}}{x}\cdot\lim_{x \to 0}\frac{\sin{x}}{1 + \cos{x}} \\ &= 1\cdot 0 \\ &= 0 \end{align*}$

**khamaar** · June 30th 2011, 07:06 AM

thank you very much, case closed.

Thanks again...Hugs!!

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