# epsilon-delta proof

• Mar 15th 2010, 01:01 PM
featherbox
epsilon-delta proof
$\frac{x+1}{x-2}\rightarrow\frac{1}{4}$ as $x\rightarrow -2$

I've got that $\frac{3}{4}\mid\frac{x+2}{x-2}\mid<\epsilon$

and obviously $x\rightarrow-2$ implies $0< x+2<\delta$
(modulus signs around x+2 in the last bit)

But the problem is getting the modulus of 1/(x-2) between something to get a value for delta in terms of epsilon.

Any help greatly appreciated.
• Mar 15th 2010, 01:35 PM
Plato
Quote:

Originally Posted by featherbox
$\frac{x+1}{x-2}\rightarrow\frac{1}{4}$ as $x\rightarrow -2$

I've got that $\frac{3}{4}\mid\frac{x+2}{x-2}\mid<\epsilon$

and obviously $x\rightarrow-2$ implies $0< x+2<\delta$
(modulus signs around x+2 in the last bit)

But the problem is getting the modulus of 1/(x-2) between something to get a value for delta in terms of epsilon.

Start with $|x+2|<1$ then you can get $\left|\frac{1}{2-x}\right|<\frac{1}{3}$.

Let $\delta =\min\left\{1,\epsilon\right\}$.