Prove Limit is sqrt{ a }

Calculus
Section 2.4

Can you do 37 in step by step fashion?

 

37.

prove that lim(sqrt(x))=sqrt(a), x->a if a>0
hint: use |sqrt(x)-sqrt(a)|=(|x-a|)/(sqrt(x)+sqrt(a))

You have to discuss the inequality |sqrt(x)−sqrt(a)|<ε.

Now take δ=min(ε*sqrt(a),a) and assume |x−a|<δ.

Since sqrt(x)+sqrt(a)>sqrt(a) , we have
|sqrt(x)−sqrt(a)|=(|x-a|)/(sqrt(x)+sqrt(a))< (|x−a|)/(sqrt(a)<δ/sqrt(a)≤ ε

 

This problem looks so scary but you did it with your eyes closed. I'm not there yet.

 

first make sure you learn definition of limits of functions , their properties, then you can start with proofs

 

I simply wanted your reply to each proof question as study notes should I ever decide to learn this stuff not needed to learn Calculus.

 

use this

 

and study notes should be in same order

 

I am doing the best I can outside of a formal classroom setting. I left calcworkshop.com a few days ago.