Limit of square root function

  • Thread starter StarTiger
  • Start date
  • #1
9
1
Could someone help me with this? I feel it should likely be easy, but I'm baffled anyway:

Homework Statement



Prove the validity of the limit lim x -> x_0 sqrt(x) = sqrt(x_0)


Homework Equations



use definition of limit


The Attempt at a Solution



Generally confused. I start with |sqrtx - sqrtx0| < epsilon
 

Answers and Replies

  • #2
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You want to show that for any delta > 0, there is an epsilon so that |sqrt(x) - sqrt(x_0)| < eps when |x - x_0| < delta.

|sqrt(x) - sqrt(x_0)| = |sqrt(x) - sqrt(x_0)| *|sqrt(x) + sqrt(x_0)| /|sqrt(x) + sqrt(x_0)| = |x - x_0|/ *|sqrt(x) + sqrt(x_0)|

Without loss of generality, you can assume that delta is reasonably small, say less than 1. That puts x within 1 unit of x_0.

Can you work with |sqrt(x) + sqrt(x_0)| to find max and min values for it?
 
  • #3
9
1
Okay...this is starting to make more sense now. Thanks for the tips. When I get it I'll try to upload what I have.
 

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