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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.

## Want to join the conversation?

• what is the need of limit?
• simply you are finding what happens to the given expression as x--> approaches a certain constant. This is widely applied in engineering, economics, astronomy and everywhere. Your imagination is the limit!
• 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!
• 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)
• 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.
• (1/0) - (1/0)?
Is it 0/0 or not possible to subtract?
• 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