# Limit

• Jan 5th 2008, 08:06 AM
clapaucius
Limit
Could anyone solve this:

lim (x -> o) ((1-cos3x) / 7x^2)

Thanks!
• Jan 5th 2008, 08:34 AM
Krizalid
Consider $\displaystyle \lim_{u \to 0} \frac{{1 - \cos u}} {{u^2 }} = \frac{1} {2}.$

Substitute $\displaystyle u=3x,$

$\displaystyle \lim_{x \to 0} \frac{{1 - \cos 3x}} {{9x^2 }} = \frac{1} {2}.$

Multiply both sides by $\displaystyle \frac97$ to get the answer.
• Jan 5th 2008, 09:36 AM
Soroban
Hello, clapaucius!

Here's one way . . . there may be better ways.

Quote:

$\displaystyle \lim_{x\to0} \frac{1- \cos3x}{7x^2}$

Multiply by $\displaystyle \frac{1+\cos3x}{1+\cos3x}$

. . $\displaystyle \frac{1-\cos3x}{7x^2}\cdot\frac{1+\cos3x}{1+\cos3x} \;=\;\frac{1-\cos^2\!3x}{7x^2(1+\cos3x)} \;=\;\frac{\sin^2\!3x}{7x^2(1 + \cos3x)}$

Multiply by $\displaystyle \frac{9}{9}$

. . $\displaystyle \frac{9}{9}\cdot\frac{\sin^2\!3x}{7x^2(1+\cos3x)} \;=\;\frac{9}{7(1+\cos3x)}\cdot\frac{\sin^2\!3x}{9 x^2} \;=\;\frac{9}{7(1+\cos3x)}\cdot\left(\frac{\sin3x} {3x}\right)^2$

Hence: . $\displaystyle \lim_{x\to0}\,\left[\frac{9}{7(1+\cos3x)}\cdot\left(\frac{\sin3x}{3x}\ right)^2\right] \;=\;\frac{9}{7(1 + \cos0)}\cdot(1^2) \;=\;\frac{9}{14}$

• Jan 5th 2008, 12:16 PM
clapaucius
$\displaystyle \frac{9}{14}$ is correct ... Thanks!!
• Jan 5th 2008, 07:04 PM
ThePerfectHacker
Use a Taylor 3rd Degree polynomial.
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