find limit as x approaches 0 of (1-cos(x))/x

[Maddy_Math]

\(\displaystyle \lim \limits_{x \to 0} \frac{1 - \cos(x)}{x} \) can I find this limit without using L' Hospital's Rule, if yes the how?

 

[HallsofIvy]

\(\displaystyle \lim \limits_{x \to 0} \frac{1 - \cos(x)}{x} \) can I find this limit without using L' Hospital's Rule, if yes the how?

Multiply both numerator and denominator by 1+ cos(x):

\(\displaystyle \dfrac{1- cos^2(x)}{x(1+ cos(x))} = \dfrac{sin^2(x)}{x(1+ cos(x))} = \left( \dfrac{sin(x)}{x} \right) \left( \dfrac{sin(x)}{1} \right) \left( \dfrac{1}{1+ cos(x)} \right)\).

 

[Maddy_Math]

Multiply both numerator and denominator by 1+ cos(x):

\(\displaystyle \dfrac{1- cos^2(x)}{x(1+ cos(x))} = \dfrac{sin^2(x)}{x(1+ cos(x))} = \left( \dfrac{sin(x)}{x} \right) \left( \dfrac{sin(x)}{1} \right) \left( \dfrac{1}{1+ cos(x)} \right)\).

and it will be (1) (0) (1/2) when we will apply limit right

Thanks for help