# How do you find the limit of (1-cosx)/(2x^2 - x) as x approaches 0?

Feb 15, 2017

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

#### Explanation:

We can use the Maclaurin series for $\cos$...

cos x = 1/(0!)-x^2/(2!)+x^4/(4!)-x^6/(6!)+...

So:

$1 - \cos x = \frac{1}{2} {x}^{2} - O \left({x}^{4}\right)$

and:

${\lim}_{x \to 0} \frac{1 - \cos x}{2 {x}^{2} - x} = {\lim}_{x \to 0} \frac{\frac{1}{2} {x}^{2} - O \left({x}^{4}\right)}{\left(2 x - 1\right) x}$

$\textcolor{w h i t e}{{\lim}_{x \to 0} \frac{1 - \cos x}{2 {x}^{2} - x}} = {\lim}_{x \to 0} \frac{\frac{1}{2} x - O \left({x}^{3}\right)}{2 x - 1}$

$\textcolor{w h i t e}{{\lim}_{x \to 0} \frac{1 - \cos x}{2 {x}^{2} - x}} = \frac{0}{- 1}$

$\textcolor{w h i t e}{{\lim}_{x \to 0} \frac{1 - \cos x}{2 {x}^{2} - x}} = 0$

Feb 15, 2017

If you've learned them, use the fundamental trigonometric limits.

#### Explanation:

${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$ (By an argument based on some geometry and the squeeze theorem.)

${\lim}_{x \rightarrow 0} \frac{1 - \cos x}{x} = 0$ (Use the limit above and trig identities.)

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

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

$= 0 \cdot \frac{1}{-} 1 = 0$