Question

Has to do with the epsilon-delta definition of a limit. Give me a second to type.

Answer 1

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

Answer 2

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

Answer 3

\[|\frac{1}{x-1}-1|<.01\]
=>
\[-.01<\frac{1}{x-1}-1<.01\]
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
\[\frac{100}{101}-1<x-2<\frac{100}{99}-1\]
\[\frac{100-101}{101}<x-2<\frac{100-99}{99}\]
\[\frac{-1}{101}<x-2<\frac{1}{99}\]

Answer 4

Remember we want
\[0<|x-2|<\delta\]

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

\[|x-2|<\delta =>- \delta <x-2<\delta \]

Answer 10

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

Answer 11

so the 1/99 doesn't matter then?

Answer 12

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

Answer 13

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