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| &lt; \delta \leq 1. Now you need <br /> \left| \frac{x - 1}{x - 3 } \right| &lt; \frac{\delta}{|x - 3|} &lt; \epsilon. You can ensure that by finding a constant K &gt; 0 such that \frac{\delta}{|x - 3|} &lt; K\delta when |x - 1| &lt; \delta \leq 1, and insisting that K\delta &lt; \epsilon.

 

[verty]

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