Question

prove the statement using the delta, epsilon definition of limits:

lim x->1 (2+4x)/3 = 2

Answer 1

Really, in this case the function does nothing nasty at x=1, so just insert x=1 into the function.

Answer 2

Hint: just choose delta to be like 10*epsilon and you'll be fine

Answer 3

you job is to show that given \(\epsilon>0\) there is some \(\delta>0\) which you write in terms of \(\epsilon\) so that if \(|x-1|<\delta \) then \(|\frac{2+4x}{3}-2|<\epsilon\)

Answer 4

a little algebra gives
\[|\frac{4(x-1)}{3}|<\epsilon\]
or
\[\frac{4}{3}|x-1|<\epsilon\] since you have control over \(|x-1|\) you can make sure that \(|x-1|<\frac{3}{4}\epsilon\) that that will do it