Solving Equation w/ L'Hopital's Rule When x = 0

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Homework Statement



I have an equation in which a term equals zero and in this case the whole equations equals zero. I know it is possible to use the L'Hopital's rule to calculate the equation but having I'm having a little trouble.

Homework Equations



The equation I would like to is the following:

k = \frac{pA}{x}*[(\frac{V}{V-Ax})^y - 1]

I have to solve this when x = 0;


The Attempt at a Solution



Plugging in 0 for x, results in 0, which should not be the answer. I figure it is possible to use the L'Hopitals rule, but currently only my denominator has the x term which would be approaching zero. But in order to use the rule I would have to manipulate the equation for both the numerator and denominator to approach zero?
 
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Instead of using de L'Hospital's rule you can (equivalently) expand the expression in the square brackets in powers of x. This should help!
 
But if I would still like to use L'Hopital's rule, how should I proceed?
 
Differentiate the numerator wrt x

Plugging in 0 for x, results in 0
Not true. It results in 0/0 !
 
Fluidman117 said:
But if I would still like to use L'Hopital's rule, how should I proceed?

Your function is ##f(x)## is of the form
f(x) = \frac{N(x)}{x}, \;\; N(x) = pA\left[ \left(\frac{V}{V-Ax}\right)^y -1\right]
When you use l'Hospital's rule (not l'Hopital!) you compute
\lim_{x \to 0} f(x) = \frac{N'(0)}{1},
so all you are really doing is just taking the first term of the series expansion of ##N(x)##. l'Hospital and series expansion are really the same thing.
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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