Prove Limit is sqrt{ a }

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

  1. nycmathguy

    nycmathguy

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    Calculus
    Section 2.4

    Can you do 37 in step by step fashion? Screenshot_20220510-193913_Samsung Notes.jpg
     
    nycmathguy, May 12, 2022
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  2. nycmathguy

    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

    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
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  4. nycmathguy

    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

    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
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  6. nycmathguy

    MathLover1

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    use this
     

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    MathLover1, May 12, 2022
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  7. nycmathguy

    MathLover1

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    and study notes should be in same order
     
    MathLover1, May 13, 2022
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  8. nycmathguy

    nycmathguy

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    I am doing the best I can outside of a formal classroom setting. I left calcworkshop.com a few days ago.
     
    nycmathguy, May 13, 2022
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