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January 30, 2015

Posted by **DJ** on Tuesday, October 4, 2011 at 10:06am.

hint: 0< |sqrt(x)-sqrt(c)|= (|x-c|)/(sqrt(x)+sqrt(c))<(1/(sqrt(c))*|x-c|

- Calculus -
**Count Iblis**, Tuesday, October 4, 2011 at 2:29pmlim x->c (sqrt(x)) = (sqrt(c))

means that the closer x is chosen to c the closer sqrt(x) comes to sqrt(c). To prove that, you need to show that making |sqrt(x) - sqrt(c)| smaller than some arbitrary epsilon is always possible by restricting x to some interval around c. Since epsilon can be any number larger than zero, this means that this interval around zero exists no matter how small epsilon is chosen. So, even if epsilon = 0.000000000001, there is stll a small interval around c for which the square root function is within 0.000000000001 of c.

From the inequalities:

0< |sqrt(x)-sqrt(c)|= (|x-c|)/(sqrt(x)+sqrt(c))<(1/(sqrt(c))*|x-c|

you can see that to require

|sqrt(x)-sqrt(c)| < epsilon

it is enough to require that:

|x-c| < sqrt(c) epsilon

So, we see that for arbitrary

epsilon > 0 there indeed does exist a delta > 0 such that for |x-c| < delta we have that

|sqrt(x) - sqrt(c)| < epsilon

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