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Epsilon-Delta Proof

  • Thread starter tvguide123
  • Start date
  • #1

Homework Statement


Let a rep. any real number greater than 0
Prove that the limit as x->a of sqrt(x) = sqrt(a)

I hav to prove the above equation using using an Epsilon-Delta proof but im not sure how to start it off.

2. The attempt at a solution

I assumed that if 0<|x-a|<d
then |f(x) - f(a)|
= |sqrt(x) - sqrt(a)|

I am allowed to use basic manipulations of numbers that preserved the equation and also make helper assumption values for delta if needed as long as i account for them in my proof.

I've been stuck on this question for 3-1/2 hours now so I would really appreciate any help!
 

Answers and Replies

  • #2
daniel_i_l
Gold Member
867
0
Try multiplying |sqrt(x) - sqrt(a)| by |sqrt(x) + sqrt(a)| /|sqrt(x) + sqrt(a)|
 
  • #3
489
0
Given epsilon>0, let delta = epsilon*sqrt(a), and remember that sqrt(x) + sqrt(a) >= sqrt(a) if x>=0.
 
  • #4
Ah thx a bunch guys, I couldnt figure out the first step for so long!

cheers :)
 

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