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L'Hôpital's rule: challenging problem

Sal uses L'Hôpital's rule to find the limit at 1 of x/(x-1)-1/lnx. Created by Sal Khan.

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• Why isn't 1/0 -1/0 directly the same as 0/0? I mean aren't they basically fractions with the same denominator, and hence addable? Or, is the fact that dividing by zero gives an undefined quantity preventing one from completing the operation?
• You nailed it here: "Or, is the fact that dividing by zero gives an undefined quantity preventing one from completing the operation?" That is the reason.

Math is a system of logic. The moment you introduce a nonsensical statement, you can prove anything, even that 1+1=0. That is why we must be very careful at each step of our work.
1/0 is nonsensical, so you can't use it as a basis of an argument, such as "since 0 is a common denominator".
• When finding the first derivative of this equation, how did Sal get by without using the quotient rule for the entire equation? Or the product rule(after converting it appropriately) on the entire equation?
• You have to take the derivatives of both the unctions independently. In the quotient rule f(x)/g(x) would be considered an entire function, changing the actual answer.
• So, wait at he got rid of a 1? How can he just remove a 1 that he was adding?
• It was +1 - 1 so he just canceled them out
• Why can´t we use L´Hôpital´s rule in indeterminate forms such as ∞-∞ ,+∞-∞, ∞+∞, ∞/0...(every other different than 0/0 and ∞/∞)?
• If you can convert the other forms into 0/0 and ∞/∞ forms (as is often done), AND after the conversion, your functions satisfy the other conditions regarding the application of L´Hôpital´s rule. then you can (indirectly) solve them!
• can L'Hopital's rule be applied as many times as we want ??
• Yes. Keep applying until you no longer get 0/0 or inf/inf.
• Isn`t it much more easier and fast @ not to use the product rule for 1/x (x-1), but instead to re-form it into just 1/x by opening the brackets and then just to derive this, which results in -x to the second, which is also one and the answer is still the same... or this is wrong by some reason?
• Yep, that's totally valid. It's faster your way.

For those who don't understand his question, he's asking why Sal didn't distribute the 1/x into the (x-1), because that would make the derivation much easier.
(1 vote)
• At , how to get one over x? I don't know what he was doing here... Is it derivative of log x equals to one over x?
• Yes, the derivative of log u = du / u
(using natural logs, of course).
Note that you have to use the chain rule, so you must multiply (not divide) by the derivative of whatever you are taking the log of. For example:

Thus the derivative of log (7x²) = d(7x²) /(7x²) = 14x/(7x²) dx = 2/x dx
• At , Sal took the limit of an expression. It came out to 1/0. Fine. Then he simplified that expression and got 0/0. Weren't those two expressions equivalent to each other? If so, why wouldn't the limits be the same? By the same token, if the limits are different, doesn't that show that the expressions aren't really the same?
• It has to do with the idea of taking the LIMIT. The functions had different exact values at the number itself. The first was not even defined at that value (0/0). However, the functions still have the same LIMIT even though some of the functions do not exist at the exact point itself.
• why couldn't he separate the limit and then solve
lim(fx-gx) should equal lim(fx)-lim(gx)
and then you could solve individually using l"hospitals rule
and subtract
is it longer then what he's doing or does it not work that way
I tried it and don't get the right answer so either i derived wrong or theres something wrong with my logic just not sure which or why