# Limit problem with square root

1. Aug 22, 2015

### chimath35

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)?

2. Aug 22, 2015

### tommyxu3

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 $(a+b)(a-b)=a^2-b^2.$

3. Aug 22, 2015

### Rinzler09

They are not equal. What was the original question?

4. Aug 22, 2015

### chimath35

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

5. Aug 22, 2015

### chimath35

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

6. Aug 22, 2015

### chimath35

also a>0

7. Aug 22, 2015

### chimath35

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)

8. Aug 22, 2015

### chimath35

never mind I figured out, but thanks for the replies

9. Aug 22, 2015

### tommyxu3

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 $|f(x)-f(a)|<\varepsilon.$

10. Aug 22, 2015

### chimath35

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

11. Aug 22, 2015

### chimath35

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.