## anonymous 5 years ago Has to do with the epsilon-delta definition of a limit. Give me a second to type.

1. anonymous

Given $f(x) = 1/ (x-1)$ Find $\delta$ such that if $0<\left| x-2 \right|<\delta$ then$\left| f(x)-1 \right|<.01$

2. anonymous

I'd appreciate it if you could walk me through how to do it.

3. myininaya

$|\frac{1}{x-1}-1|<.01$ => $-.01<\frac{1}{x-1}-1<.01$ Add 1 to all sides $1-.01<\frac{1}{x-1}<1+.01$ $.99<\frac{1}{x-1}<1.01$ $\frac{99}{100}< \frac{1}{x-1}<\frac{101}{100}$ $\frac{100}{101}<x-1<\frac{100}{99}$ Subtract 1 on all sides $\frac{100}{101}-1<x-2<\frac{100}{99}-1$ $\frac{100-101}{101}<x-2<\frac{100-99}{99}$ $\frac{-1}{101}<x-2<\frac{1}{99}$

4. myininaya

Remember we want $0<|x-2|<\delta$

5. anonymous

Yeah, I think I can take it from here. Just reading through it again :)

6. anonymous

wait, do we need the absolute values and the zero, then? How do you get those?

7. anonymous

x−2 > - 1/101 then |x−2| > 1/101 and |x−2|<δ ...... therefore δ > 1/101

8. myininaya

$\delta=\frac{1}{101}$ I chose delta to be the smallest

9. myininaya

$|x-2|<\delta =>- \delta <x-2<\delta$

10. myininaya

so since we want the smallest value for delta I chose delta=1/101

11. anonymous

so the 1/99 doesn't matter then?

12. myininaya

no you just want the smallest possible value for delta you can find

13. anonymous

oh, okay. we kinda skipped through this part in class months ago. hard to review but i think i'm getting the gist of it