# 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

#### Fantini

Poirot, that is a good idea. However, since he doesn't wish to use l'Hôpital's rule (or perhaps can't, therefore he wishes not to), I'd say that using power series is probably out of context as well. Nevertheless, another clever suggestion! We already have four possible paths to the solution. 