# How do you find the limit of (1-cosx)/(xsinx) as x->0?

Nov 12, 2017

${\lim}_{x \rightarrow 0} \frac{1 - \cos x}{x \sin x} = \frac{1}{2}$

#### Explanation:

First of all, since as $x \rightarrow 0$, $\sin x \rightarrow 0$ also, we can rewrite the denominator as ${x}^{2}$.

Hence we need to find: ${\lim}_{x \rightarrow 0} \frac{1 - \cos x}{{x}^{2}}$

Since this still results in an indeterminate $\frac{0}{0}$, we apply L'Hopital's Rule.

$\frac{\frac{d}{\mathrm{dx}} \left(1 - \cos x\right)}{\frac{d}{\mathrm{dx}} \left({x}^{2}\right)} = \sin \frac{x}{2 x}$

If we substitute 'approaching zero' as a less formal $\frac{1}{\infty}$, we arrive at the expression: $\frac{\frac{1}{\infty}}{\frac{2}{\infty}}$.

After cancelling out the infinities, this leaves $\frac{1}{2}$.

Or simply let $\sin x = x$ again, which gives:

$\sin \frac{x}{2 x} = \frac{x}{2 x} = \frac{1}{2}$.

Nov 12, 2017

Use ${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$

#### Explanation:

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

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

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

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

$= \left(1\right) \left(1\right) \left(\frac{1}{1 + 1}\right) = \frac{1}{2}$