limit as x goes to infinity question involving square roots

[km707]

Hi there, this question has been bugging me for a long time and I'm not sure what to do with it.

limit as x -> infinity:

x[sup:2ld5p4h8]3/2[/sup:2ld5p4h8][(1+x+1/x)[sup:2ld5p4h8]1/2[/sup:2ld5p4h8] - (1+x)[sup:2ld5p4h8]1/2[/sup:2ld5p4h8]]

The answer should be about 1/2 if I graphed the equation correctly.

1st failed method:
I isolated the x terms in each of the square roots to obtain:

x[sup:2ld5p4h8]3/2[/sup:2ld5p4h8][x[sup:2ld5p4h8]1/2[/sup:2ld5p4h8]*(1/x + 1 + 1/x[sup:2ld5p4h8]2[/sup:2ld5p4h8])[sup:2ld5p4h8]1/2[/sup:2ld5p4h8] - x[sup:2ld5p4h8]1/2[/sup:2ld5p4h8]*(1/x +1)[sup:2ld5p4h8]1/2[/sup:2ld5p4h8]]

= x[sup:2ld5p4h8]2[/sup:2ld5p4h8][(1/x + 1 + 1/x[sup:2ld5p4h8]2[/sup:2ld5p4h8])[sup:2ld5p4h8]1/2[/sup:2ld5p4h8] - (1/x +1)[sup:2ld5p4h8]1/2[/sup:2ld5p4h8]]

... and that leaves me stuck as I don't know what to do with the 1st square root with 3 terms in it - the 2nd one I would usually binomially expand the expression and cancel terms away..

2nd failed method:

multiplied x[sup:2ld5p4h8]3/2[/sup:2ld5p4h8] through by bringing x[sup:2ld5p4h8]3[/sup:2ld5p4h8]into the square rooted terms. ...and I'm still pretty much stuck with the same situation.

I have no idea what to do with the expression with 3 terms inside a square root.

HELP! I won't be able to sleep until I know!!

Thanks!

 

[soroban]

Hello, km707!

This is a strange one . . . I'll get you started . . .

\(\displaystyle \lim_{x\to\infty}\bigg[x^{\frac{3}{2}}\left(\sqrt{1 + x + \tfrac{1}{x}} - \sqrt{1+x}\right)\bigg]\)

Multiply top and bottom by the "conjugate":

. . \(\displaystyle \frac{x^{\frac{3}{2}}\bigg(\sqrt{1+x+\tfrac{1}{x}} - \sqrt{1+x}\bigg)}{1}\cdot\frac{\sqrt{1+x+\frac{1}{x}} + \sqrt{1+x}}{\sqrt{1+x+\frac{1}{x}} + \sqrt{1+x}}\)

. . \(\displaystyle = \;\frac{x^{\frac{3}{2}}\bigg[\left(1 + x+ \frac{1}{x}\right) - (1 + x)\bigg]}{\sqrt{1+x+\frac{1}{x}} + \sqrt{1+x}} \;=\;\frac{x^{\frac{3}{2}}\left(\frac{1}{x}\right)}{\sqrt{1+x+\frac{1}{x}} + \sqrt{1+x}}\)

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

Can you continue?

 

[km707]

I think so..

x[sup:20orlzon]1/2[/sup:20orlzon]/ [x[sup:20orlzon]1/2[/sup:20orlzon][(1/x + 1 + 1/x[sup:20orlzon]2[/sup:20orlzon])[sup:20orlzon]1/2[/sup:20orlzon] + (1 + 1/x)[sup:20orlzon]1/2[/sup:20orlzon]]

The x[sup:20orlzon]1/2[/sup:20orlzon] on the top and bottom cancel, and 1/x and 1/x[sup:20orlzon]2[/sup:20orlzon] go to zero as x goes to infinity so we are left with 1/(1+1) = 1/2!!

Thank you very much! I find this topic is very badly taught, but I suppose its all about knowing how to use mathematical tricks and shortcuts!