# Epsilon delta proof

• Sep 5th 2010, 11:58 PM
Steel88
Epsilon delta proof
$$\lim_{x \to 0} \frac{3x}{x^2-4x} = -\frac{3}{4} \ \ \epsilon = 1.5$
$

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

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.

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

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

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

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.
• Sep 6th 2010, 12:32 AM
Prove It
You need to show

$0 < \left|x - 0\right| < \delta \implies \left|\frac{3x}{x^2 - 4x} - \left(- \frac{3}{4}\right)\right| < \epsilon$.

Solve the inequality $\left|\frac{3x}{x^2 - 4x} - \left(-\frac{3}{4}\right)\right| < \epsilon$ for $|x - 0| = |x|$

$\left|\frac{3x}{x^2 - 4x} - \left(-\frac{3}{4}\right)\right| < \epsilon$

$\left|\frac{3x}{x^2 - 4x} + \frac{3}{4}\right| < \epsilon$

$\left|\frac{12x}{4(x^2 - 4x)} + \frac{3(x^2 - 4x)}{4(x^2 - 4x)}\right| < \epsilon$

$\left|\frac{12x + 3(x^2 - 4x)}{4(x^2 - 4x)}\right|<\epsilon$

$\left|\frac{3x^2}{4x(x - 4)}\right| < \epsilon$

$\left|\frac{3x}{4(x - 4)}\right| < \epsilon$

$\frac{|x|}{4|x - 4|} < \frac{\epsilon}{3}$.

So we want to find a number $M$ such that $\frac{1}{4|x - 4|} \leq M$.

$M|x| < \frac{\epsilon}{3}$

$|x| < \frac{\epsilon}{3M}$, i.e. $\delta = \frac{\epsilon}{3M}$

If we let $|x| < 5$ then

$-5 < x < 5$

$-9 < x - 4 < 1$.

So $|x - 4| < 1$

$4|x-4| < 4$

$\frac{1}{4|x-4|} < \frac{1}{4}$.

So set $M = \frac{1}{4}$ and define $\delta = \min\left\{\frac{\epsilon}{3M}, 5 \right \}$

$= \min\left\{\frac{\epsilon}{3\left(\frac{1}{4}\righ t)}, 5 \right\}$

$= \min\left\{\frac{4\epsilon}{3}, 5\right\}$

and we are set to write the proof.

Proof:

Let $\epsilon > 0$ and define $\delta = \min\left\{\frac{4\epsilon}{3}, 5 \right\}$. Then if $0 < |x| < \delta$ we have

$\left|\frac{3x}{x^2 - 4x} - \left(-\frac{3}{4}\right)\right| = \left|\frac{12x}{4(x^2 - 4x)} + \frac{3(x^2 - 4x)}{4(x^2 - 4x)}\right|$

$= \left|\frac{12x + 3x^2 - 12x}{4x(x - 4)}\right|$

$= \left|\frac{3x}{4(x - 4)}\right|$

$= \frac{3|x|}{4|x - 4|}$

$< \frac{3|x|}{4}$ since $\frac{1}{4|x - 4|} < \frac{1}{4}$

$< \frac{3\left(\frac{4\epislon}{3}\right)}{4}$ since $|x| < \delta = \frac{4\epsilon}{3}$

$< \epsilon$.

Therefore $0 < |x - 0| < \delta \implies \left|\frac{3x^2}{x^2 - 4x} - \left(-\frac{3}{4}\right)\right| < \epsilon$, and thus

$\lim_{x \to 0}\frac{3x^2}{x^2 - 4x} = -\frac{3}{4}$.
• Sep 6th 2010, 08:15 AM
Steel88
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

$
\frac{|x|}{4|x - 4|} < \frac{\epsilon}{3}
$

Is this where I can plug in numbers for $\epsilon$ 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 $|x| < \delta$

Outside that, why did you pick 5 for "If we let $|x| < 5$"

And I don't even come close to understand the $\delta = \min\left\{\frac{\epsilon}{3M}, 5 \right \}$
• Sep 6th 2010, 09:04 AM
Quote:

Originally Posted by Steel88
Is this where I can plug in numbers for $\epsilon$ 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 $|x| < \delta$

You can't "plug in" epsilon. $\epsilon > 0$. The "game" (a poor one, admittedly), is that for any epsilon I choose, you can choose a delta such that |f(x) - f(a)| < $\epsilon$

Quote:

Outside that, why did you pick 5 for "If we let $|x| < 5$"
Quote:

If we let $|x| < 5$ then

$-5 < x < 5$

$-9 < x - 4 < 1$.

So $|x - 4| < 1$

$4|x-4| < 4$

$\frac{1}{4|x-4|} < \frac{1}{4}$.
This inequality is later used in the proof. edit;// we can do this because we can make delta as small as we please.

Quote:

And I don't even come close to understand the $\delta = \min\left\{\frac{\epsilon}{3M}, 5 \right \}$
Suppose I choose a epsilon. If you choose $\delta = 5$, and |f(x)-f(a)| < $\epsilon$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)
• Sep 6th 2010, 09:57 AM
Steel88
Okay, however I when I choose epsilon as 1.5, I end up with $\frac{|x|}{4|x - 4|} < \frac{\ 1.5}{3}$ Which gives me .5 = delta, however I'm still left with $\frac{|x|}{4|x - 4|}$ On the left, or if you substitute M its $M|x| < \frac{\epsilon}{3}$ or $\delta = \frac{\epsilon}{3M}$ 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.
• Sep 6th 2010, 05:35 PM
Prove It