1. ## Epsilon delta proof

$\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 $signs, couldn't get the <, > signs up without it popping up. 2. You need to show$\displaystyle 0 < \left|x - 0\right| < \delta \implies \left|\frac{3x}{x^2 - 4x} - \left(- \frac{3}{4}\right)\right| < \epsilon$. Solve the inequality$\displaystyle \left|\frac{3x}{x^2 - 4x} - \left(-\frac{3}{4}\right)\right| < \epsilon$for$\displaystyle |x - 0| = |x|\displaystyle \left|\frac{3x}{x^2 - 4x} - \left(-\frac{3}{4}\right)\right| < \epsilon\displaystyle \left|\frac{3x}{x^2 - 4x} + \frac{3}{4}\right| < \epsilon\displaystyle \left|\frac{12x}{4(x^2 - 4x)} + \frac{3(x^2 - 4x)}{4(x^2 - 4x)}\right| < \epsilon\displaystyle \left|\frac{12x + 3(x^2 - 4x)}{4(x^2 - 4x)}\right|<\epsilon\displaystyle \left|\frac{3x^2}{4x(x - 4)}\right| < \epsilon\displaystyle \left|\frac{3x}{4(x - 4)}\right| < \epsilon\displaystyle \frac{|x|}{4|x - 4|} < \frac{\epsilon}{3}$. So we want to find a number$\displaystyle M$such that$\displaystyle \frac{1}{4|x - 4|} \leq M$.$\displaystyle M|x| < \frac{\epsilon}{3}\displaystyle |x| < \frac{\epsilon}{3M}$, i.e.$\displaystyle \delta = \frac{\epsilon}{3M}$If we let$\displaystyle |x| < 5$then$\displaystyle -5 < x < 5\displaystyle -9 < x - 4 < 1$. So$\displaystyle |x - 4| < 1\displaystyle 4|x-4| < 4\displaystyle \frac{1}{4|x-4|} < \frac{1}{4}$. So set$\displaystyle M = \frac{1}{4}$and define$\displaystyle \delta = \min\left\{\frac{\epsilon}{3M}, 5 \right \}\displaystyle = \min\left\{\frac{\epsilon}{3\left(\frac{1}{4}\righ t)}, 5 \right\}\displaystyle = \min\left\{\frac{4\epsilon}{3}, 5\right\}$and we are set to write the proof. Proof: Let$\displaystyle \epsilon > 0$and define$\displaystyle \delta = \min\left\{\frac{4\epsilon}{3}, 5 \right\} $. Then if$\displaystyle 0 < |x| < \delta$we have$\displaystyle \left|\frac{3x}{x^2 - 4x} - \left(-\frac{3}{4}\right)\right| = \left|\frac{12x}{4(x^2 - 4x)} + \frac{3(x^2 - 4x)}{4(x^2 - 4x)}\right|\displaystyle = \left|\frac{12x + 3x^2 - 12x}{4x(x - 4)}\right|\displaystyle = \left|\frac{3x}{4(x - 4)}\right|\displaystyle = \frac{3|x|}{4|x - 4|}\displaystyle < \frac{3|x|}{4}$since$\displaystyle \frac{1}{4|x - 4|} < \frac{1}{4}\displaystyle < \frac{3\left(\frac{4\epislon}{3}\right)}{4}$since$\displaystyle |x| < \delta = \frac{4\epsilon}{3}\displaystyle < \epsilon$. Therefore$\displaystyle 0 < |x - 0| < \delta \implies \left|\frac{3x^2}{x^2 - 4x} - \left(-\frac{3}{4}\right)\right| < \epsilon$, and thus$\displaystyle \lim_{x \to 0}\frac{3x^2}{x^2 - 4x} = -\frac{3}{4}$. 3. Great response thank you I've spent probably 5 hours on this over the weekend. However I'm still a little confused on a few parts. When we're at this point$\displaystyle
\frac{|x|}{4|x - 4|} < \frac{\epsilon}{3}
$Is this where I can plug in numbers for$\displaystyle \epsilon $because this is the only point where I see that I can plug in 1.5 for epsilon, and get .5 in return for delta. If this is the case do I only need to solve the numerator on this side to equal$\displaystyle |x| < \delta $Outside that, why did you pick 5 for "If we let$\displaystyle |x| < 5$" And I don't even come close to understand the$\displaystyle \delta = \min\left\{\frac{\epsilon}{3M}, 5 \right \}$4. Originally Posted by Steel88 Is this where I can plug in numbers for$\displaystyle \epsilon $because this is the only point where I see that I can plug in 1.5 for epsilon, and get .5 in return for delta. If this is the case do I only need to solve the numerator on this side to equal$\displaystyle |x| < \delta $You can't "plug in" epsilon.$\displaystyle \epsilon > 0 $. The "game" (a poor one, admittedly), is that for any epsilon I choose, you can choose a delta such that |f(x) - f(a)| <$\displaystyle \epsilon $Outside that, why did you pick 5 for "If we let$\displaystyle |x| < 5$" If we let$\displaystyle |x| < 5$then$\displaystyle -5 < x < 5\displaystyle -9 < x - 4 < 1$. So$\displaystyle |x - 4| < 1\displaystyle 4|x-4| < 4\displaystyle \frac{1}{4|x-4|} < \frac{1}{4}$. This inequality is later used in the proof. edit;// we can do this because we can make delta as small as we please. And I don't even come close to understand the$\displaystyle \delta = \min\left\{\frac{\epsilon}{3M}, 5 \right \}$Suppose I choose a epsilon. If you choose$\displaystyle \delta = 5$, and |f(x)-f(a)| <$\displaystyle \epsilon$you are done. If not, you need to choose something lower (hence the min). In order to choose this delta without fail, (this is a game you want to win), then it must be calculated based on what epsilon I choose. (which in this case comes from the work established above) 5. Okay, however I when I choose epsilon as 1.5, I end up with$\displaystyle \frac{|x|}{4|x - 4|} < \frac{\ 1.5}{3} $Which gives me .5 = delta, however I'm still left with$\displaystyle \frac{|x|}{4|x - 4|}$On the left, or if you substitute M its$\displaystyle M|x| < \frac{\epsilon}{3}$or$\displaystyle \delta = \frac{\epsilon}{3M} \$ Am I just supposed to ignore the denominator if I have |X| in the numerator, or just ignore the M otherwise?

I'm just not seeing how when I choose the epsilon value where I get delta in return...In every other problem I've done with this the left side just collapses to equal the right and gives you the delta.

6. I suggest you download and read the PDF attached in the following link, as they explain it much better than I could...

http://www.mathhelpforum.com/math-he...ofs-47767.html