Delta-Epsilon Proof: Prove lim_{x\implies 1} \frac{2}{x-3} = -1

[knowLittle]

Homework Statement

Prove that
$lim_{x\implies 1} \frac{2}{x-3} = -1$

Use delta-epsilon.

The Attempt at a Solution

Proof strategy:
$| { \frac{ 2}{x-3} +1 } | < \epsilon$$\frac{x-1}{x-3} < \epsilon$
, since delta have to be a function of epsilon alone and not include x. I need to restrict delta
$|x-1 | < 1 \leq \delta \\ -3 < x-3 < -1$

I know that there's something wrong. Help?

What if I say that
$-2 = x -3 \\ 1 =x \\ \frac{ 1-1}{-2} < \epsilon \\ \delta=min(1, \epsilon)$

Does it make any sense?

 

[pasmith]

Homework Statement

Prove that
$lim_{x\implies 1} \frac{2}{x-3} = -1$

Use delta-epsilon.

The Attempt at a Solution

Proof strategy:
$| { \frac{ 2}{x-3} +1 } | < \epsilon$$\frac{x-1}{x-3} < \epsilon$
, since delta have to be a function of epsilon alone and not include x. I need to restrict delta
$|x-1 | < 1 \leq \delta \\ -3 < x-3 < -1$

I know that there's something wrong. Help?

You don't know how small $\delta$ may need to be, but you can decide that it's not going to be bigger than 1. That gives you $$|x - 1| < \delta \leq 1.$$ Now you need $$\left| \frac{x - 1}{x - 3 } \right| < \frac{\delta}{|x - 3|} < \epsilon.$$ You can ensure that by finding a constant $K > 0$ such that $$\frac{\delta}{|x - 3|} < K\delta$$ when $|x - 1| < \delta \leq 1$, and insisting that $K\delta < \epsilon$.

 

[verty]

You're almost there, now prove that your choice for delta works.