Prove that the lim as x goes to a of sqrtx = sqrta

  • Thread starter apiwowar
  • Start date
can i get some help on proving this limit?
 

lurflurf

Homework Helper
2,417
122
start with
sqrt(x)-sqrt(a)=(x-a)/(sqrt(x)+sqrt(a))
 
I think it would be very helpful if you guys used latex

The question should be

Proove that [tex]\lim {x \rightarrow a} \sqrt{x} = \sqrt{a}[/tex]

And the first reply should be

start with
[tex]\sqrt{x}-\sqrt{a}=\frac{x-a}{\sqrt{x}+\sqrt{a}}[/tex]

For basic things like this, latex only takes about a minute to learn and it VERY much easier to read :)
 

HallsofIvy

Science Advisor
Homework Helper
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857
In order to prove that "[itex]\lim_{x\to a}\sqrt{x}= \sqrt{a}[/itex], you need to prove that "Given [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]|x- a|< \delta[/itex] then [itex]|\sqrt{x}-\sqrt{a}|< \epsilon[/itex]".

Start from the second inequality and show (using the above hint) that you can find such a [itex]\delta[/itex].
 

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