Question

Prove the statement using ɛ, ɓ definition of a limit.

Lim (1+(1/3)x)=2
x->3

Answer 1

definition in this case would be
given any \(\epsilon>0\) there is a \(\delta\) which you write as a function of \(\epsilon\) that if \(|x-3|<\delta\) then \(|1+\frac{1}{3}x-2|<\epsilon\)

Answer 2

work backwards
\[|\frac{1}{3}x-1|<\epsilon\]\[-\epsilon<\frac{1}{3}x-1<\epsilon\]\[1-\epsilon<\frac{1}{3}x<\epsilon+1\]
\[3-3\epsilon<x<3\epsilon+3\]

Answer 3

the last one is the same as
\[|x-3|<3\epsilon\] to you can say "let \(\delta=3\epsilon\)" work the algebra backwards and you have your proof

Answer 4

not really a coincident that your line with slope \(m\) requires a \(\delta\) of \(\frac{1}{m}\)

Answer 5

Is there any easy way to do/explain this cause I don't really understand the steps that you are doing.

Answer 6

well ok we need to show that if |x-3| < delta then |f(x)-L|<epsilon for any given epsilon
so he worked backwards which is always a great way to start these, and started where we need to end up, he showed that if that statement is true, then |x-3|<3epsilon
then the proof is done
by working backwards we got to |x-3|<3e must hold, and that is nice because that is exactly what we need to assume
so
assume
let epsilon be given
if delta = 3epsilopn
then\[|x-3|<\delta=3\epsilon\\implies\\|\frac{1}{3}x-1|=\frac{|x-3|}{|3|}\le\epsilon\]

Answer 7

i dont think I made anything clearer but I gave it a shot

Answer 8

Just wondering where he got the \[|x-3| < \delta\]

Answer 9

\[|\frac{1}{3}x-1|<\epsilon\\|\frac{x-3}{3}|<\epsilon\\\frac{|x-3|}{|3|}<|x-3|<\epsilon\]

Answer 10

\[so\space given\space \epsilon\\let\space\delta=\epsilon\\if\space|x-3|<\delta\\then\\|1+\frac{1}{3}x-2|<\frac{|x-3|}{3}<|x-3| <\epsilon\]