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Limit calculation

Yankel

Active member
Jan 27, 2012
398
Hello,

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

[tex]\frac{1-cos(x)}{x^{2}}[/tex]

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
Jan 26, 2012
4,192
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

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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? (Smile)
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hello,

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

[tex]\frac{1-cos(x)}{x^{2}}[/tex]

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

Administrator
Staff member
Feb 24, 2012
13,775
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
Feb 15, 2012
250
cosx=1-x^2/2+O(x^4) as x -> 0
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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! (Clapping)

We already have four possible paths to the solution. (Dance)