# How To Find The Limits Algebraically...?

• September 24th 2008, 08:34 PM
AlphaRock
How To Find The Limits Algebraically...?
lim ((sin^2)x)/x
x->0

This is a tricky question... Can anybody lend a helping hand?
• September 24th 2008, 08:36 PM
Chris L T521
Quote:

Originally Posted by AlphaRock
lim ((sin^2)x)/x
x->0

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

$\lim_{x\to{0}}\frac{\sin^2(x)}{x}=\lim_{x\to{0}}\f rac{\sin(x)}{x}\cdot \sin(x)=\dots$

You should know what $\lim_{x\to{0}}\frac{\sin(x)}{x}$ equals.

Does this make sense?

--Chris
• September 25th 2008, 10:34 PM
AlphaRock
Quote:

Originally Posted by Chris L T521
$\lim_{x\to{0}}\frac{\sin^2(x)}{x}=\lim_{x\to{0}}\f rac{\sin(x)}{x}\cdot \sin(x)=\dots$

You should know what $\lim_{x\to{0}}\frac{\sin(x)}{x}$ equals.

Does this make sense?

--Chris

Thanks, Chris.

What I can't understand is why

lim ((sin^2)x)/x
x->0

will equal...

lim (sinx)/x times sin(x)
x->0
• September 25th 2008, 10:39 PM
kirbyiwaki
$\lim_{x\to{0}}\frac{\sin^2(x)}{x}=\lim_{x\to{0}}\f rac{\sin(x)}{x}\cdot \sin(x)$

Just factorize the

$\sin^2(x)$

into

$\sin(x)\cdot\sin(x)$
• September 25th 2008, 10:39 PM
Moo
Quote:

Originally Posted by AlphaRock
Thanks, Chris.

What I can't understand is why

lim ((sin^2)x)/x
x->0

will equal...

lim (sinx)/x times sin(x)
x->0

Because $\sin^2(x)=[\sin(x)]^2=\sin(x) \times \sin(x)$
• October 6th 2008, 06:11 PM
AlphaRock
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
• October 6th 2008, 06:19 PM
11rdc11
Quote:

Originally Posted by AlphaRock
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

Is this what it looks like?

$\lim_{x \to 0} \frac{\frac{1}{2+x} - \frac{1}{2}}{x}$
• October 6th 2008, 06:30 PM
AlphaRock
Quote:

Originally Posted by 11rdc11
Is this what it looks like?

$\lim_{x \to 0} \frac{\frac{1}{2+x} - \frac{1}{2}}{x}$

Yup!
• October 6th 2008, 06:39 PM
11rdc11
$\lim_{x \to 0} \frac{\frac{1}{2+x} - \frac{1}{2}}{x}$

Find common denominator in the numerator

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

which equals

$\lim_{x \to 0}~\frac{-1}{4+2x}~=~\frac{-1}{4}$
• October 15th 2008, 08:47 PM
AlphaRock
Quote:

Originally Posted by 11rdc11
$\lim_{x \to 0} \frac{\frac{1}{2+x} - \frac{1}{2}}{x}$

Find common denominator in the numerator

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

which equals

$\lim_{x \to 0}~\frac{-1}{4+2x}~=~\frac{-1}{4}$

Can somebody explain this?

$\frac{\frac{-x}{4+2x}}{x} = \frac{-x}{x(4+2x)}$

Because I'm having troubles understanding how you got there...
• October 15th 2008, 08:49 PM
Chris L T521
Quote:

Originally Posted by AlphaRock
Can somebody explain this?

$\frac{\frac{-x}{4+2x}}{x} = \frac{-x}{x(4+2x)}$

Because I'm having troubles understanding how you got there...

$\frac{\frac{-x}{4+2x}}{x} =\frac{\frac{-x}{4+2x}}{1}\cdot\frac{1}{x} = \frac{-x}{4+2x}\cdot\frac{1}{x} =\frac{-x}{x(4+2x)}$

Does this make sense?

--Chris