∞ - ∞ = L'H but how does this become a fraction?

• Mar 23rd 2009, 05:21 AM
TYTY
∞ - ∞ = L'H but how does this become a fraction?
$
\sqrt{n^2 - 3n} - n
$

This is to get the limit as n approaches ∞

In the book this became a fraction somehow. There is probably some algebraic property that I am oblivious of. My prediction is that in 30 minutes I will refresh this page and say "OMG so obvious" (Wink)
• Mar 23rd 2009, 05:25 AM
mr fantastic
Quote:

Originally Posted by TYTY
$
\sqrt{n^2 - 3n} - n
$

This is to get the limit as n approaches ∞

In the book this became a fraction somehow. There is probably some algebraic property that I am oblivious of. My prediction is that in 30 minutes I will refresh this page and say "OMG so obvious" (Wink)

$\sqrt{n^2 - 3n} - n = \frac{(\sqrt{n^2 - 3n} - n) \cdot (\sqrt{n^2 - 3n} + n)}{\sqrt{n^2 - 3n} + n} = \frac{(n^2 - 3n) - n^2}{\sqrt{n^2 - 3n} + n} = \frac{-3n}{\sqrt{n^2 - 3n} + n}$

No need for L'H if you now divide numerator and denominator by n: $\frac{-3}{\sqrt{1 - \frac{3}{n}} + 1}$.

Now take the limit.

Let me save you the trouble ...... "OMG so obvious"
• Mar 23rd 2009, 05:32 AM
TYTY
Quote:

Originally Posted by mr fantastic
$\sqrt{n^2 - 3n} - n = \frac{(\sqrt{n^2 - 3n} - n) \cdot (\sqrt{n^2 - 3n} + n)}{\sqrt{n^2 - 3n} + n} = \frac{(n^2 - 3n) - n^2}{\sqrt{n^2 - 3n} + n} = \frac{-3n}{\sqrt{n^2 - 3n} + n}$

No need for L'H if you now divide numerator and denominator by n: $\frac{-3}{\sqrt{1 - \frac{3}{n}} + 1}$.

Now take the limit.

Let me save you the trouble ...... "OMG so obvious"

Thank you mister fantastic; that answer looks very good. What I'm still not entirely sure of was what compelled you to multiply by

$
{\sqrt{n^2 - 3n} + n}/{\sqrt{n^2 - 3n} + n}
$

I mean, obviously it is mathematically sound and allowed you to get the answer but I'm just not following the train of thought that had you multiplying by that exact ratio in the first place.
• Mar 23rd 2009, 05:37 AM
mr fantastic
Quote:

Originally Posted by TYTY
Thank you mister fantastic; that answer looks very good. What I'm still not entirely sure of was what compelled you to multiply by

$
{\sqrt{n^2 - 3n} + n}/{\sqrt{n^2 - 3n} + n}
$

I mean, obviously it is mathematically sound and allowed you to get the answer but I'm just not following the train of thought that had you multiplying by that exact ratio in the first place.

The simplest answer is to say that it's a trick that works, learned from experience.
• Mar 23rd 2009, 05:44 AM
TYTY
So would you say that when you see ∞ - ∞, one of the things you try first is (∞ - ∞)[(∞ + ∞)/(∞ + ∞)]? Did I understand correctly?

Sorry for no latex as I'm late for work...
• Mar 23rd 2009, 08:21 AM
HallsofIvy
What mr. fantastic was doing was "rationalizing the numerator".