Limit Calc: 1-cos(x)/x^2 w/o L'Hôpital's Rule

In summary, Poirot suggests that the problem may be best solved by using a power series. Furthermore, he suggests that this method might be better suited for a different context.
  • #1
Yankel
395
0
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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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)
 
  • #4
Yankel said:
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

$\chi$ $\sigma$
 
  • #5
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:

$\displaystyle \lim_{x\to0}\frac{1-\cos(x)}{x^2}$
 
  • #6
cosx=1-x^2/2+O(x^4) as x -> 0
 
  • #7
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)
 

FAQ: Limit Calc: 1-cos(x)/x^2 w/o L'Hôpital's Rule

1. What is the formula for calculating the limit of 1-cos(x)/x^2 without using L'Hôpital's Rule?

The formula for calculating this limit is lim x→0 (1-cos(x))/x^2 = 1/2.

2. Why can't L'Hôpital's Rule be used to find this limit?

L'Hôpital's Rule can only be used when the limit is in an indeterminate form, such as 0/0 or ∞/∞. In this case, the limit is already defined and does not fall under an indeterminate form.

3. How do you prove that the limit of 1-cos(x)/x^2 without using L'Hôpital's Rule is 1/2?

This limit can be proved using the trigonometric identity lim x→0 (1-cos(x))/x^2 = lim x→0 2sin^2(x)/x^2. Then, by using the basic limit lim x→0 sin(x)/x = 1, we can rewrite the equation as lim x→0 2sin(x)/x * lim x→0 sin(x)/x = 2*1*1 = 2. Finally, we divide by 2 to get the desired result of 1/2.

4. Can you use other methods to find the limit of 1-cos(x)/x^2 without L'Hôpital's Rule?

Yes, there are other methods that can be used to find this limit, such as using Maclaurin series or the squeeze theorem. However, the simplest and most efficient method is using the trigonometric identity as mentioned in the previous answer.

5. What is the significance of the limit of 1-cos(x)/x^2 without L'Hôpital's Rule?

This limit is significant because it is a fundamental limit in calculus and is often used as a building block for more complex limit calculations. It also has important applications in physics and engineering, particularly in the study of oscillatory motion.

Similar threads

Replies
1
Views
1K
Replies
5
Views
2K
Replies
4
Views
2K
Replies
3
Views
1K
Replies
3
Views
1K
Replies
2
Views
1K
Back
Top