# Limit

• January 27th 2008, 07:09 PM
Limit
$\lim_{x\to 0}\frac{1-cos\;x}{x^2}$ without l'Hopital Rule.

Thank you
• January 27th 2008, 08:50 PM
mr fantastic
Quote:

$\lim_{x\to 0}\frac{1-cos\;x}{x^2}$ without l'Hopital Rule.

Thank you

Substitute the Maclaurin Series for cos x, simplify and take the limit:

$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - ......$

Therefore you have $\frac{1 - 1 + \frac{x^2}{2} - \frac{x^4}{24} + .....}{x^2} = \frac{1}{2} - \frac{x^2}{24} + .....$.

As x --> 0 the limiting value is obvious .....
• January 27th 2008, 08:54 PM
ThePerfectHacker
Quote:

$\lim_{x\to 0}\frac{1-cos\;x}{x^2}$ without l'Hopital Rule.

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

Now use the fact that $\lim_{x\to 0}\frac{\sin x}{x} = 1$.

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Using infinite series is like shooting a dead fly with a Grand Piano.
• January 27th 2008, 09:26 PM
mr fantastic
Quote:

Originally Posted by ThePerfectHacker
[tex][snip]
Using infinite series is like shooting a dead fly with a Grand Piano.

True.

But I've found most students can be taught to kill even a live fly using a Grand Piano. As opposed to teaching them how to pull a rabbit out of a hat (even when it's a fairly ordinary rabbit, as in the present case) .....

By the way, I don't mean to be critical of the rabbit - I love magic - it's just that most students struggle to learn it. Even when they do, they usually pull out a dead fish .....

My view, all comment welcome.
• January 28th 2008, 04:34 AM
curvature
Quote:

$\lim_{x\to 0}\frac{1-cos\;x}{x^2}$ without l'Hopital Rule.