# Evaluate lim_(x->0)(1-cosx)/x^2?

## Thanks!

Dec 18, 2017

$\frac{1}{2}$

#### Explanation:

L'Hopital's rules says that the ${\lim}_{x \to a} \frac{f \left(x\right)}{g \left(x\right)} \implies \frac{f ' \left(a\right)}{g ' \left(a\right)}$

Using this, we get ${\lim}_{x \to 0} \frac{1 - \cos x}{x} ^ 2 \implies \frac{- \sin 0}{2 \left(0\right)}$

Yet as the denominator is $0$, this is impossible. So we do a second limit:
$\lim \left(x \to 0\right) \frac{\sin x}{2 x} \implies \frac{\cos 0}{2} = \frac{1}{2} = 0.5$

So, in total
${\lim}_{x \to 0} \frac{1 - \cos x}{x} ^ 2 \implies {\lim}_{x \to 0} \frac{\sin x}{2 x} \implies \cos \frac{x}{2} \implies \cos \frac{0}{2} = \frac{1}{2}$

Dec 18, 2017

${\lim}_{x \rightarrow 0} \frac{1 - \cos x}{x} ^ 2 = {\lim}_{x \rightarrow 0} \left({\sin}^{2} \frac{x}{x} ^ 2 \cdot \frac{1}{1 + \cos x}\right) = \frac{1}{2}$

#### Explanation:

$\frac{1 - \cos x}{x} ^ 2 = \frac{\left(1 - \cos x\right)}{x} ^ 2 \cdot \frac{\left(1 + \cos x\right)}{\left(1 + \cos x\right)}$

 = (1-cos^2x)/(x^2(1+cosx)

 = sin^2x/(x^2(1+cosx)

$= {\sin}^{2} \frac{x}{x} ^ 2 \cdot \frac{1}{1 + \cos x}$

Aug 10, 2018

$\frac{1}{2}$

#### Explanation:

We see that through direct evaluation, we get indeterminate form, $\frac{0}{0}$. Next, we can use L'Hôpital's Rule, which says

${\lim}_{x \to a} \frac{f \left(x\right)}{g} \left(x\right) = {\lim}_{x \to a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

We can take the derivative of our numerator and denominator function to obtain the new limit

${\lim}_{x \to 0} \frac{\sin x}{2 x}$

Through direct evaluation, we would get indeterminate form again, so we can take the derivatives once more to get

${\lim}_{x \to 0} \frac{\cos x}{2}$

When we evaluate this limit at zero, we get

${\lim}_{x \to 0} \frac{1 - \cos x}{{x}^{2}} = \frac{1}{2}$

Hope this helps!