# Limit calculation

#### Yankel

##### Active member
Hello,

I need some assistance with this limit when x->0:

$$\frac{1-cos(x)}{x^{2}}$$

I do not wish to use L'Hôpital's rule.

What I tried was to write x^2 as x*x, and to separate it so I can use the known limit of cos(x)/x, however it only brought me to the 0*infinity problem.

#### Ackbach

##### Indicium Physicus
Staff member
Do you not want to use l'Hôpital's rule because the probem given to you wants you to use a different method? That is, must you not use it? Or would you just personally prefer a different way? l'Hôpital's rule solves this problem in two lines.

#### Fantini

MHB Math Helper
Not using l'Hôpital's rule is just as fast. Multiply and divide by $1 + \cos x$, yielding

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

Can you finish from here?

#### chisigma

##### Well-known member
Hello,

I need some assistance with this limit when x->0:

$$\frac{1-cos(x)}{x^{2}}$$

I do not wish to use L'Hôpital's rule.

What I tried was to write x^2 as x*x, and to separate it so I can use the known limit of cos(x)/x, however it only brought me to the 0*infinity problem.
A possibility in this particular case is to use the identity...

$\displaystyle 1 - cos x = 2\ \sin^{2} \frac{x}{2}$ (1)

... and from (1) to derive...

$\displaystyle \frac{1 - \cos x}{x^{2}} = \frac{1}{2}\ \frac{\sin^{2} \frac{x}{2}}{(\frac{x}{2})^{2}}$ (2)

In the right term of (2) there is a 'fundamental limit' and this limit is 1...

Kind regards

$\chi$ $\sigma$

#### MarkFL

Staff member
I came late to the party! (d)

All I have to add is that to write the limit with $\displaystyle \LaTeX$, use the following code:

\lim_{x\to0}\frac{1-\cos(x)}{x^2}

to get:

$\displaystyle \lim_{x\to0}\frac{1-\cos(x)}{x^2}$

#### Poirot

##### Banned
cosx=1-x^2/2+O(x^4) as x -> 0