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.
Taking the Limit, as x -> A, of (1/x - 1/A) / (x - A), where A > 0
- Thread starter tnick03
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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.
\(\displaystyle \displaystyle \lim _{x\to A}\left(\frac{\frac{1}{x}-\frac{1}{A}}{x-A}\right)\)
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.
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Steven G
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Is this not just the derivative of f(x) = 1/x at x=A?
So f ' (x) =.... and f ' (A) =....
Steven G
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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
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\(\displaystyle \displaystyle {\left(\dfrac{\frac{1}{x}-\frac{1}{A}}{x-A}\right)=\left(\dfrac{\frac{A-x}{Ax}}{x-A}\right)=-\dfrac{1}{Ax}}\)