# Thread: How To Find The Limits Algebraically...?

1. ## How To Find The Limits Algebraically...?

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

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

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

This is a tricky question... Can anybody lend a helping hand?
$\displaystyle \lim_{x\to{0}}\frac{\sin^2(x)}{x}=\lim_{x\to{0}}\f rac{\sin(x)}{x}\cdot \sin(x)=\dots$

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

Does this make sense?

--Chris

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

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

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

Just factorize the

$\displaystyle \sin^2(x)$

into

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

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

6. 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):

7. 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?

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

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

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

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

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

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

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

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

$\displaystyle \lim_{x \to 0}~\frac{-1}{4+2x}~=~\frac{-1}{4}$
Can somebody explain this?

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

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

11. Originally Posted by AlphaRock
Can somebody explain this?

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

Because I'm having troubles understanding how you got there...
$\displaystyle \frac{\frac{-x}{4+2x}}{x} =\frac{\frac{-x}{4+2x}}{1}\cdot\frac{1}{x} = \frac{-x}{4+2x}\cdot\frac{1}{x} =\frac{-x}{x(4+2x)}$

Does this make sense?

--Chris