# Prove Limit is sqrt{ a }

Discussion in 'Calculus' started by nycmathguy, May 12, 2022.

1. ### nycmathguy

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Calculus
Section 2.4

Can you do 37 in step by step fashion?

nycmathguy, May 12, 2022

2. ### MathLover1

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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

MathLover1, May 12, 2022
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3. ### nycmathguy

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

nycmathguy, May 12, 2022
4. ### MathLover1

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

MathLover1, May 12, 2022
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5. ### nycmathguy

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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.

nycmathguy, May 12, 2022
6. ### MathLover1

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use this

#### Attached Files:

• ###### Calculus_Cheat_Sheet_Limits.pdf
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116.2 KB
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MathLover1, May 12, 2022
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7. ### MathLover1

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and study notes should be in same order

MathLover1, May 13, 2022
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