Help Prove Using Epsilon-Delta Limits
H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,763 Nov 5, 2020 #2 "limx→af(x)=L\displaystyle \lim_{x\to a} f(x)= Lx→alimf(x)=L" means "Given ϵ>0\displaystyle \epsilon> 0ϵ>0 there exist δ>0\displaystyle \delta> 0δ>0 such that if ∣x−a∣<δ\displaystyle |x- a|< \delta∣x−a∣<δ then ∣f(x)−L∣<ϵ\displaystyle |f(x)- L|< \epsilon∣f(x)−L∣<ϵ". Here, f(x)= x and L= a so that becomes "Given ϵ>0\displaystyle \epsilon> 0ϵ>0 there exist δ>0\displaystyle \delta> 0δ>0 such that if ∣x−a∣<δ\displaystyle |x- a|< \delta∣x−a∣<δ then ∣x−a∣<ϵ\displaystyle |x- a|< \epsilon∣x−a∣<ϵ". Comparing the two inequalities you should see they are the same- you just have to take δ=ϵ\displaystyle \delta= \epsilonδ=ϵ. Typically that observation is enough but if you wanted to give a "rigorous" proof you would say "Given ϵ>0\displaystyle \epsilon> 0ϵ>0 if ∣x−a∣<ϵ\displaystyle |x- a|< \epsilon∣x−a∣<ϵ then ∣f(x)−L∣=∣x−a∣<ϵ\displaystyle |f(x)- L|= |x- a|< \epsilon∣f(x)−L∣=∣x−a∣<ϵ".
"limx→af(x)=L\displaystyle \lim_{x\to a} f(x)= Lx→alimf(x)=L" means "Given ϵ>0\displaystyle \epsilon> 0ϵ>0 there exist δ>0\displaystyle \delta> 0δ>0 such that if ∣x−a∣<δ\displaystyle |x- a|< \delta∣x−a∣<δ then ∣f(x)−L∣<ϵ\displaystyle |f(x)- L|< \epsilon∣f(x)−L∣<ϵ". Here, f(x)= x and L= a so that becomes "Given ϵ>0\displaystyle \epsilon> 0ϵ>0 there exist δ>0\displaystyle \delta> 0δ>0 such that if ∣x−a∣<δ\displaystyle |x- a|< \delta∣x−a∣<δ then ∣x−a∣<ϵ\displaystyle |x- a|< \epsilon∣x−a∣<ϵ". Comparing the two inequalities you should see they are the same- you just have to take δ=ϵ\displaystyle \delta= \epsilonδ=ϵ. Typically that observation is enough but if you wanted to give a "rigorous" proof you would say "Given ϵ>0\displaystyle \epsilon> 0ϵ>0 if ∣x−a∣<ϵ\displaystyle |x- a|< \epsilon∣x−a∣<ϵ then ∣f(x)−L∣=∣x−a∣<ϵ\displaystyle |f(x)- L|= |x- a|< \epsilon∣f(x)−L∣=∣x−a∣<ϵ".
