lim ((sin^2)x)/x

x->0

This is a tricky question... Can anybody lend a helping hand?

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- Sep 24th 2008, 08:34 PMAlphaRockHow To Find The Limits Algebraically...?
lim ((sin^2)x)/x

x->0

This is a tricky question... Can anybody lend a helping hand? - Sep 24th 2008, 08:36 PMChris L T521
- Sep 25th 2008, 10:34 PMAlphaRock
- Sep 25th 2008, 10:39 PMkirbyiwaki
$\displaystyle \lim_{x\to{0}}\frac{\sin^2(x)}{x}=\lim_{x\to{0}}\f rac{\sin(x)}{x}\cdot \sin(x)$

Just factorize the

$\displaystyle \sin^2(x)$

into

$\displaystyle \sin(x)\cdot\sin(x)$ - Sep 25th 2008, 10:39 PMMoo
- Oct 6th 2008, 06:11 PMAlphaRock
Ah! I understand why now...

Thanks, guys.

I'm having trouble with this new limit question:

lim ((1/(2+x)) - (1/2))/x

x->0

P.S. How do you guys format stuff like this (I just copy and pasted):

http://www.mathhelpforum.com/math-he...a4fa953d-1.gif - Oct 6th 2008, 06:19 PM11rdc11
- Oct 6th 2008, 06:30 PMAlphaRock
- Oct 6th 2008, 06:39 PM11rdc11
$\displaystyle \lim_{x \to 0} \frac{\frac{1}{2+x} - \frac{1}{2}}{x}$

Find common denominator in the numerator

$\displaystyle \lim_{x \to 0} \frac{\frac{2-2-x}{2(2+x)}}{x} = \frac{\frac{-x}{4+2x}}{x} = \frac{-x}{x(4+2x)}$

which equals

$\displaystyle \lim_{x \to 0}~\frac{-1}{4+2x}~=~\frac{-1}{4}$ - Oct 15th 2008, 08:47 PMAlphaRock
- Oct 15th 2008, 08:49 PMChris L T521