1. ## epsilon-delta proof

$\displaystyle \frac{x+1}{x-2}\rightarrow\frac{1}{4}$ as$\displaystyle x\rightarrow -2$

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

and obviously $\displaystyle x\rightarrow-2$ implies $\displaystyle 0< x+2<\delta$
(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.

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

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

and obviously $\displaystyle x\rightarrow-2$ implies $\displaystyle 0< x+2<\delta$
(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 $\displaystyle |x+2|<1$ then you can get $\displaystyle \left|\frac{1}{2-x}\right|<\frac{1}{3}$.

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