# Limit problem with square root

I have done this problem before but forgot how to get from one step to the next:

let a>0.

how is absval(x^1/2-a^1/2) equal to abval(x-a)/(x^1/2-a^1/2)?

I think you want to express
$$\sqrt{x}-\sqrt{a}=\frac{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}{\sqrt{x}+\sqrt{a}}=\frac{x-a}{\sqrt{x}+\sqrt{a}}$$
right? If so, it just applies the law ##(a+b)(a-b)=a^2-b^2.##

chimath35
They are not equal. What was the original question?

I think you want to express
$$\sqrt{x}-\sqrt{a}=\frac{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}{\sqrt{x}+\sqrt{a}}=\frac{x-a}{\sqrt{x}+\sqrt{a}}$$
right? If so, it just applies the law ##(a+b)(a-b)=a^2-b^2.##
but the abs val goes away on the denom. that is what I don't understand, you can't assume root x - root a is positive

They are not equal. What was the original question?
let f(x) equal root x show that as x→a lim f(x) equals root a

let f(x) equal root x show that as x→a lim f(x) equals root a
also a>0

but the abs val goes away on the denom. that is what I don't understand, you can't assume root x - root a is positive
sorry I meant you can't assume root x plus root a is positive, if say c∈ℝ and c>0 then abs val(c⋅x)=c⋅abs val(x)

never mind I figured out, but thanks for the replies

To show ##\displaystyle{\lim_{x\rightarrow a}}\sqrt{x}=\sqrt{a},## you ought to prove for any ##x\in (a-\delta, a+\delta),## ##\exists \varepsilon## such that ##|f(x)-f(a)|<\varepsilon.##

To show ##\displaystyle{\lim_{x\rightarrow a}}\sqrt{x}=\sqrt{a},## you ought to prove for any ##x\in (a-\delta, a+\delta),## ##\exists \varepsilon## such that ##|f(x)-f(a)|<\varepsilon.##
Thanks, yeah I have done this proof before. I just made a mistake thinking about negatives when clearly root x plus root a is always positive and therefore can be pulled out of the abs val

Thanks, yeah I have done this proof before. I just made a mistake thinking about negatives when clearly root x plus root a is always positive and therefore can be pulled out of the abs val
I mean it assumes that x>0. The problem did not state that although I guess it is just assumed considering it is real analysis.