Question

A limit proof exercise:

Using the limit definition proof that

lim[(x+1)/x]=3/2 as x-->2

Answer 1

you mean with \(\epsilon\) and \(\delta\) and all that?

Answer 2

yep

Answer 3

just prove that lim(x+1)=3 as x approaches 2
and that limx = 2 as x approaches 2, and then use my awesome proof (that's the long way)
Satellite will give u the short way in a few seconds.

Answer 4

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

Answer 5

http://openstudy.com/study#/updates/4fe5ec2ee4b02c91101aceba
Proof that the limit of a quotient is the quotient of the limits.

Answer 6

if you want a snap proof it is that your function is continuous at \(x=2\) but that is not what you want.

Answer 7

algebra in the absolute value gives
\[|\frac{2-x}{2x}|<\epsilon\]
so you want that one. that is good because you have control over the numerator
\[|2-x|=|x-2|\] so you only need worry about the denominator, but since \(x\) is close to two you can assert that \(1<x<3\) for sure and so \(\frac{|2-x|}{|2x|}\leq |x-2|\)

Answer 8

therefore i think in this case you can say "make \(\delta = \epsilon\) and for \(\epsilon<1\)" you get your result, arguing backwards from the bottom up

Answer 9

not sure if that is clear, but i believe it is right. you start with saying "given any \(\epsilon >0\) if \(|x-2|<\epsilon\) and \(\epsilon <1\) we have
\[|\frac{x+1}{x}-\frac{3}{2}|=\frac{|x-2|}{|2x|}\leq |x-2|<\epsilon\]

Answer 10

or something close to that.

Answer 11

okey, i'm going to do it on the paper, thanx satellite73!

Answer 12

check my algebra, and then make sure it is clear what the reasoning is. these tend to be annoying the first time you see them because you have to work backwards.