$\displaystyle \[

\lim_{x \to 0} \frac{3x}{x^2-4x} = -\frac{3}{4} \ \ \epsilon = 1.5

\]

$

Now, the answer I already know is $\displaystyle \delta = .5$

Ive tried at least 20 different things, but cannot figure out what I should be doing, or if I'm even allowed to do some of the things I have been doing. The book we use sucks, and barely covers this.

$\displaystyle \frac{3x}{x^2-4x}-(-\frac{3}{4}) \$< $ \ 1.5 $

Which reduces to this after putting under a common denominator and reducing an x.

$\displaystyle \frac{3x}{4x-16} \$<$ 1.5 $

At this point I can't figure out what to do, whether its 1.5 or just using epsilon.

edit: and don't mind the $ signs, couldn't get the <, > signs up without it popping up.