Taking the Limit, as x -> A, of (1/x - 1/A) / (x - A), where A > 0

[tnick03]

I am having trouble with this problem. It is lim as x approaches A of (1/x - 1/A) / (x - A) when A is a positive constant. So I plugged A in for x and both the numerator and denominator turn into zero. That would mean I could use l'hospital's rule. But if I do that then they just both remain zero and the derivative of zero is zero. Maybe I am missing something or my math is wrong but I need help.

 

[ksdhart2]

Just to clarify, the limit you're tasked with finding is this, correct?

$\displaystyle \displaystyle \lim _{x\to A}\left(\frac{\frac{1}{x}-\frac{1}{A}}{x-A}\right)$

One approach to this problem is to notice that, if we let f(x) = 1/x, this is precisely the definition of the derivative (EDIT: to be accurate, it's the derivative of 1/x at the point x = A). Therefore, we know the answer must be the derivative of 1/x. Another approach is to, as you indicated, use L'Hopital's Rule. Your post suggests there's a misunderstanding of what L'Hopital's rule says. You appear to have plugged in the value of x=A (taken the limit) and then taken the derivative. But that's not correct - you must take the derivative first. i.e.

$\displaystyle \displaystyle \lim _{x\to A}\left(\dfrac{\frac{1}{x}-\frac{1}{A}}{x-A}\right)=\lim _{x\to A}\left(\frac{\frac{d}{dx}\left(\frac{1}{x}-\frac{1}{A}\right)}{\frac{d}{dx}\left(x-A\right)}\right)$

The derivative of the sum of two functions is equal to the sum of the derivatives of those functions, so we can further simplify:

$\displaystyle \displaystyle \lim _{x\to A}\left(\frac{\frac{d}{dx}\left(\frac{1}{x}\right)-\dfrac{d}{dx}\left(\frac{1}{A}\right)}{\frac{d}{dx}\left(x\right)-\frac{d}{dx}\left(A\right)}\right)$

Now you try continuing from here.

 

[Steven G]

I am having trouble with this problem. It is lim as x approaches A of (1/x - 1/A) / (x - A) when A is a positive constant. So I plugged A in for x and both the numerator and denominator turn into zero. That would mean I could use l'hospital's rule. But if I do that then they just both remain zero and the derivative of zero is zero. Maybe I am missing something or my math is wrong but I need help.

Is this not just the derivative of f(x) = 1/x at x=A?
So f ' (x) =.... and f ' (A) =....

 

[Steven G]

One approach to this problem is to notice that, if we let f(x) = 1/x, this is precisely the definition of the derivative.

Be careful. That limit is not the definition of the derivative (I assume of f(x)). Rather it is the definition of the derivative of f(x) at x = A. :shock: (please don't hate me)

 

[pka]

I am having trouble with this problem. It is lim as x approaches A of (1/x - 1/A) / (x - A) when A is a positive constant.

$\displaystyle \displaystyle {\left(\dfrac{\frac{1}{x}-\frac{1}{A}}{x-A}\right)=\left(\dfrac{\frac{A-x}{Ax}}{x-A}\right)=-\dfrac{1}{Ax}}$