How To Find The Limits Algebraically...?

**AlphaRock** · September 24th 2008, 08:34 PM

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

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

**Chris L T521** · September 24th 2008, 08:36 PM

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

**AlphaRock** · September 25th 2008, 10:34 PM

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

**kirbyiwaki** · September 25th 2008, 10:39 PM

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

Just factorize the

$\sin^2(x)$

into

$\sin(x)\cdot\sin(x)$

**Moo** · September 25th 2008, 10:39 PM

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 $\sin^2(x)=[\sin(x)]^2=\sin(x) \times \sin(x)$

**AlphaRock** · October 6th 2008, 06:11 PM

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):

**11rdc11** · October 6th 2008, 06:19 PM

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):

Is this what it looks like?

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

**AlphaRock** · October 6th 2008, 06:30 PM

Originally Posted by 11rdc11

Is this what it looks like?

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

Yup!

**11rdc11** · October 6th 2008, 06:39 PM

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

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

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

**AlphaRock** · October 15th 2008, 08:47 PM

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...

**Chris L T521** · October 15th 2008, 08:49 PM

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

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