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Calculus/l'hospitals rule

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Expert: - 5/18/2009


Question
i know you've said i you dont get this, but i think that you can, because you have done such a good job on my previous questions, so if you can please help me i'd aprreciate it... thanks

Find A so that lim (x+A/x-2A)^x=5
              x->00

use l'hospital rule


Answer
Well, when you first asked me, I didn't think I had a clue.
Now that you have asked me again, it seems like that problem has been in the back of my head for awhile, and I think I have the answer...

I didn't use l'Hosital's rule.

What I did say was that
(1+a)^n is known as 1 + na + n(n-1)a²/2 +  ...

Using this, it can be said that (1 + A/x)^x =
 1+ xA/x + x(x-1)A²/(2x²) + ....
and that (1 - 2A/x)^x was equal to
 1 - 2Ax/x + (x(x-1)/2)(2A)²/2 -...

In this way, as x -> ∞, all of the terms divided by x disappear as
n->∞ and all that is left is ΣA^n/n! in the numerator and Σ(-2A)^n/n!
in the denominator.

The looks like e^A and the denominator looks like e^(-2A).
We have e^A/e^(-2A) = 5, which goes to e^(3A) = 5, so we know that
3A = ln(5) => A = ln(5)/3.

From what I remember from l'Hospital's rule, it involves taking the derivative of the top and bottom if the top and bottom both go to 0 or the top and bottom both go to ∞.  To me, that would seem to make it even more complicated than it is already.

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