What is the Proof for the Limit of Square Root Function?

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Homework Help Overview

The discussion revolves around proving the limit of the square root function as \( x \) approaches \( a \), specifically focusing on the condition that \( a > 0 \). Participants are exploring the formal definition of limits and how to manipulate expressions involving square roots.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to establish the limit using the epsilon-delta definition, but expresses uncertainty about how to relate delta to epsilon in their proof. Another participant suggests a manipulation of the expression involving square roots to aid in the proof.

Discussion Status

Participants are actively engaging with the problem, with one providing a potential direction for the original poster's approach. There is a sense of collaboration, but no consensus or resolution has been reached yet.

Contextual Notes

One participant mentions the late hour, indicating a potential time constraint affecting their engagement with the problem.

marco0009
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Homework Statement


Prove: \lim_{x \to a} \sqrt{x} = \sqrt{a}, if a>0


Homework Equations


|f(x)-L| < \epsilon
|x-a| < \delta


The Attempt at a Solution


|\sqrt{x}-\sqrt{a}| < \epsilon, when |x-a| < \delta
|x - 2\sqrt{x}\sqrt{a}-a| < \epsilon^2, when |x-a| < \delta

From here I don't know where to go. I don't see any obvious way to get delta into terms of epsilon.
 
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well |\sqrt{x}-\sqrt{a}| =|(\sqrt{x}-\sqrt{a})\frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}| can you see what to do now?
 
Yep. Thanks, it has been a long night.
 
marco0009 said:
Yep. Thanks, it has been a long night.
It is 3 in the morning here, lol !
 

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