- #1

- 110

- 0

let a>0.

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter chimath35
- Start date

- #1

- 110

- 0

let a>0.

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

- #2

- 240

- 42

$$\sqrt{x}-\sqrt{a}=\frac{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}{\sqrt{x}+\sqrt{a}}=\frac{x-a}{\sqrt{x}+\sqrt{a}}$$

right? If so, it just applies the law ##(a+b)(a-b)=a^2-b^2.##

- #3

- 3

- 0

They are not equal. What was the original question?

- #4

- 110

- 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

$$\sqrt{x}-\sqrt{a}=\frac{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}{\sqrt{x}+\sqrt{a}}=\frac{x-a}{\sqrt{x}+\sqrt{a}}$$

right? If so, it just applies the law ##(a+b)(a-b)=a^2-b^2.##

- #5

- 110

- 0

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

- #6

- 110

- 0

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

- #7

- 110

- 0

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)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

- #8

- 110

- 0

never mind I figured out, but thanks for the replies

- #9

- 240

- 42

- #10

- 110

- 0

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

- 110

- 0

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.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

Share: