- #1

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## 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

- Thread starter StarTiger
- Start date

- #1

- 9

- 1

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

use definition of limit

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

- #2

Mark44

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|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?

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