# [SOLVED] Limits. LIMITS!

• Feb 25th 2008, 09:31 PM
spinscent
[SOLVED] Limits. LIMITS!
lim x --> 0

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

. . . I know the answer's -1/4, I just don't know how to get there.

I'm not really even sure why it's confusing me. I just got done with a chapter about limits... Maybe the x's in both numerator and denominator are killing me..?

I know I can't substitute. Gwah!
• Feb 25th 2008, 10:41 PM
Jhevon
Quote:

Originally Posted by spinscent
lim x --> 0

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

. . . I know the answer's -1/4, I just don't know how to get there.

I'm not really even sure why it's confusing me. I just got done with a chapter about limits... Maybe the x's in both numerator and denominator are killing me..?

I know I can't substitute. Gwah!

do you mean $\displaystyle \lim_{x \to 0} \frac {\frac 1{2 + x} - \frac 12}x$? if so, use parenthese!

there are not many options here. combine the fractions in the numerator.

$\displaystyle \Rightarrow \lim_{x \to 0} \frac {\frac {2 - 2 - x}{2(2 + x)}}x = \lim_{x \to 0} \frac {-x}{2(2 + x)} \cdot \frac 1x$

now what?