lim[x->infty][sqrt(x^2+x)-sqrt(x^2-x)]/[sqrt(x^2+2x)-sqrt(x^

[green_tea]

Hi!
I need some help with this:

lim [sqrt(x^2 + x) - sqrt(x^2 - x)] /[sqrt(x^2 + 2x) - sqrt(x^2 - 2x)]
x-->?

I've been trying a lot of things, for example to multiplie the numerator with its conjugate, [sqrt(x^2 + x) + sqrt(x^2 - x)] to get rid of the sqrt's in he denominator, but when I do that the denominator gets a bit complicated. What I get is:

[sqrt(x^2 + x) - sqrt(x^2 - x)] /[sqrt(x^2 + 2x) - sqrt(x^2 - 2x)] * [sqrt(x^2 + x) + sqrt(x^2 - x)]/[sqrt(x^2 + x) + sqrt(x^2 - x)] =

= 2x / [sqrt(x^4 + 3x^3 + 2x^2) + sqrt(x^4 + x^3 -2x^2) - sqrt(x^4 - x^3 - 2x^2) - sqrt(x^4- 3x^3 + 2x^2)]

and I don't know what to do next. I tried to simplifie by "taking out" x^4 from the square roots in the denominator, so that I can get rid of the x in the numerator but when I do that I end up with

2/x*[sqrt(1 + 3/x + 2/x^2) + sqrt(1 + 1/x - 2/x^2) - sqrt(1 - 1/x - 2/x^2) - sqrt(1 - 3/x + 2/x^2)]

and here the denominator goes towards ?*(1+1-1-1)= ?*0 when x goes to ?, so it's hard to tell what happens with the whole equation...
So now I'm stuck and I don't know what to do.. Maybe I just started wrong?
Does anyone know how to solve this one? Help... :|

 

[mmm4444bot]

Re: problem with limes x-->?

Hello Green Tea:

The following may not be the easiest method, but here's my reasoning.

1) Multiply and divide the numerator by its conjugate.

2) Multiply and divide the denominator by its conjugate.

SYMBOLIC REPRESENTATION

N = original numerator

N[sub:c24h2ohv]c[/sub:c24h2ohv] = conjugate of original numerator

D = original denominator

D[sub:c24h2ohv]c[/sub:c24h2ohv] = conjugate of original denominator

\(\displaystyle \frac{\frac{N}{1}}{\frac{D}{1}} \cdot \frac{\frac{N_c}{N_c}}{\frac{D_c}{D_c}}\)

This should yield the following.

\(\displaystyle \frac{\frac{2x}{\sqrt{x^2 + x} \;+\; \sqrt{x^2 - x}}}{\frac{4x}{\sqrt{x^2 + 2x} \;+\; \sqrt{x^2 - 2x}}} \;=\; \frac{1}{2} \cdot \frac{\sqrt{x^2 + 2x} \;+\; \sqrt{x^2 - 2x}}{\sqrt{x^2 + x} \;+\; \sqrt{x^2 - x}}\)

In this last result, you can factor out x^2 from each of the four radicands.

EXAMPLE

\(\displaystyle \sqrt{x^2 + 2x} \;=\; \sqrt{x^2} \cdot \sqrt{1 + \frac{2}{x}}\)

This leads to a cancellation of all x^2 terms.

Apply the limit to what's left over, and I believe you'll get (1/2) * (1).

Cheers,

~ Mark