# L'Hoptial's rule

1. Mar 11, 2014

### Fluidman117

1. The problem statement, all variables and given/known data

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.

2. Relevant 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;

3. 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?

2. Mar 11, 2014

### vanhees71

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!

3. Mar 11, 2014

### Fluidman117

But if I would still like to use L'Hopital's rule, how should I proceed?

4. Mar 11, 2014

### BvU

Differentiate the numerator wrt x

Not true. It results in 0/0 !

5. Mar 11, 2014

### Ray Vickson

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: Mar 11, 2014