Limit w. conjugate/ L'Hopital's Rule

**Smac** · September 17th 2008, 08:16 AM

our instructor gave us this problem as a heads up because it will be on our quiz tomorrow, but it is pretty tough. He told us to multiply by the conjugate...

lim (x^2+x)^1/2 + x
x -> (infinite)

I have determined that the conjugate is (x^2+x)^1/2 - x, but when i do that, and then to L'hopital's, i feel as tho continuing will get me nowhere... help?

**Moo** · September 17th 2008, 08:47 AM

Hello,

Originally Posted by Smac

our instructor gave us this problem as a heads up because it will be on our quiz tomorrow, but it is pretty tough. He told us to multiply by the conjugate...

lim (x^2+x)^1/2 + x
x -> (infinite)

I have determined that the conjugate is (x^2+x)^1/2 - x, but when i do that, and then to L'hopital's, i feel as tho continuing will get me nowhere... help?

for the conjugate !

Now, multiply and divide by the conjugate :

$\sqrt{x^2+x}+x=(\sqrt{x^2+x}+x) \times \frac{\sqrt{x^2+x}-x}{\sqrt{x^2+x}-x}$

$=\frac{(\sqrt{x^2+x})^2-x^2}{\sqrt{x^2+x}-x}$

$=\frac{x}{\sqrt{x^2+x}-x}$

You can apply l'Hospital's, but there are nicer methods

:

$=\frac{x}{\sqrt{x^2\left(1+\frac 1x\right)}-x}=\frac{x}{x \sqrt{1+\frac 1x}-x}$

$=\frac{x}{x \left(\sqrt{1+\frac 1x}-1\right)}=\frac{1}{\sqrt{1+\frac 1x}-1}}$

LImit to infinity : $\sqrt{1+\frac 1x} \to 1$ , with superior values to 1. So $\sqrt{1+\frac 1x}-1 \to 1$ with positive values.

Thus the limit is...

(you would have found the same with l'Hospital's)

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My point is that the limit would have been more friendly if it was (x²+x)^(1/2) - x.

**Smac** · September 17th 2008, 11:08 AM

wow....that was easy as hell <----stupid

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