- #1
member 587159
Homework Statement
Proof that:
##\lim_{x \to 1} \frac{1}{1+x} = \frac{1}{2}##
using the epsilon-delta definition of a limit.
(Problem below)
Homework Equations
##\lim_{x \to 1} \frac{1}{1+x} = \frac{1}{2} \iff \forall ε>0, \existsδ>0, \forall x ∈ \mathbb{R}\backslash\{-1\}: 0 < |x-1| < δ \Rightarrow |\frac{1}{1+x} - \frac{1}{2}|<ε##
The Attempt at a Solution
Draft[/B]
So, I have to find δ(ε) such that:
##|x-1| < δ \Rightarrow |\frac{1}{1+x} - \frac{1}{2}|<ε##
and, after simplifying I got:
##|\frac{1}{1+x} - \frac{1}{2}|<ε \Rightarrow |x-1| < ε|2x+2|##
Now, I chose ##δ = 1##, then ##|x-1| < 1##, thus ##2< 2x + 2 < 6##. So I decided to take ##δ = min\{1,6ε\}##
Proof
We take ε arbitrary and we choose ##δ = min\{1,6ε\}##
Case 1: δ = 1, then 6ε > 1 as ##δ = min\{1,6ε\}##.
We then have, ##|x-1|<δ \Rightarrow |x-1| < 1 \Rightarrow |x-1| < 6ε \Rightarrow \frac{|x-1|}{6} < ε##
Now, here comes the problem. I wanted to use that ##6>|2x + 2|## (this result was obtained in the draft). However, when I would do this, the total fraction will become larger (as the denominator becomes smaller), and it would be possible that the inequality does not hold anymore. I got a similar problem with case 2. I'm learning this on my own, so it is possible that I'm totally wrong. Thank you for helping me in advance. Also, posting in Latex for the first time. Any hints on this matter would also be appreciated.
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