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

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SUMMARY

The limit of the expression (1 - cos(x)) / x² as x approaches 0 can be evaluated without using L'Hôpital's Rule by applying trigonometric identities and fundamental limits. The identity 1 - cos(x) = 2sin²(x/2) simplifies the limit to (1/2) * (sin²(x/2) / (x/2)²), which approaches 1 as x approaches 0. This method provides a clear and efficient solution to the limit problem while adhering to the constraints of avoiding L'Hôpital's Rule.

PREREQUISITES
  • Understanding of trigonometric identities, specifically 1 - cos(x) = 2sin²(x/2)
  • Familiarity with limits in calculus, particularly fundamental limits involving sin(x)
  • Basic knowledge of LaTeX for mathematical notation
  • Ability to manipulate algebraic expressions involving limits
NEXT STEPS
  • Study the derivation and application of trigonometric identities in limit evaluations
  • Learn about fundamental limits, particularly lim (x→0) (sin(x)/x) = 1
  • Explore alternative methods for solving limits without L'Hôpital's Rule
  • Practice writing mathematical expressions in LaTeX for clearer presentation
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Students and educators in calculus, mathematicians seeking alternative limit evaluation techniques, and anyone interested in deepening their understanding of trigonometric limits without relying on L'Hôpital's Rule.

Yankel
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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.
 
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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.
 
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)
 
Yankel said:
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

$\chi$ $\sigma$
 
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:

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

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