Prove Limit is sqrt{ x }

[nycmathguy]

Calculus
Section 1.7

Number 37 is not your typical classroom limit prove question but it looks interesting.

 

[MathLover1]

proof:

We assume that the limit is taken within the domain of definition of f(x)=sqrt(x). Then, give a>0

|sqrt(x) -qrt(0)|<a whenever, 0<x<δ=a

if given a>0, we first restrict δ<a/2, then, with x element of [a/2,3a/2] we have for any given a>0

|x -a|/ |sqrt(x) +qrt(a)|

<= |x -a|/ |sqrt(a/2) +qrt(a)|

<=(2abs(a - x))/((2 + sqrt(2)) sqrt(a))
< a

whenever, |x-a|<δ=min(a/2,(1+2sqrt(2))*a)

 

[nycmathguy]

proof:

We assume that the limit is taken within the domain of definition of f(x)=sqrt(x). Then, give a>0

|sqrt(x) -qrt(0)|<a whenever, 0<x<δ=a

if given a>0, we first restrict δ<a/2, then, with x element of [a/2,3a/2] we have for any given a>0

|x -a|/ |sqrt(x) +qrt(a)|

<= |x -a|/ |sqrt(a/2) +qrt(a)|

<=(2abs(a - x))/((2 + sqrt(2)) sqrt(a))
< a

whenever, |x-a|<δ=min(a/2,(1+2sqrt(2))*a)

This is really complicated, fuzzy material. More time is needed to review your reply.