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Current time:0:00Total duration:7:50

Video transcript

we want to figure out the limit the limit as X approaches 1 of the expression x over X minus 1 minus 1 over the natural log of X so let's just see what happens when we just try to plug in the 1 what happens if we evaluate this expression at 1 well then we're going to get we're going to get a 1 here over 1 minus 1 so we're going to get something like a 1 over a 0 minus 1 over and what's the natural log of 1 e to the what power is equal to 1 well either the or anything to the 0 power is equal to 1 so e to the 0th power is going to be equal to 1 so the natural log of 1 is going to be 0 so we get the strange undefined 1 over 0 2 minus 1 over 0 it's this bizarre-looking undefined form but it's not the indeterminate type of form that we looked for in law petals rule we're not getting a 0 over 0 we're not getting an infinity over infinity so you might just say hey ok this is a non la petals rule problem we're going to figure out this limit some other way and I would say well you know don't don't give up just yet maybe we can manipulate this algebraically somehow so that it will give us the lappet all indeterminate form and then we can just apply the rule and to do that let's just see if what happens if we add these two expressions so if we add them so this expression let's if we add it it'll be well the common denominator is going to be X minus 1 X minus 1 times the natural log of X I just multiplied the denominators and then the numerator is going to be it's going to be well if I I multiply essentially this whole term by natural log of X so it's going to be X natural log of X X natural log of X and then this whole term I'm going to multiply it by X minus 1 so minus X minus 1 so minus X minus 1 and you could break it apart and see that this expression in this expression of the same thing this right here that right there is the same thing as x over X minus 1 because the natural log of X is canceled out we get rid of that then this right here this right here is the same thing as one over natural log of X because the X minus ones cancel out so hopefully you realize all I did is I added these two expressions so given that let's see what happens if I take the limit as X approaches 1 of this thing because these are the same thing do we get anything more interesting so what do we have here we have 1 times the natural log of 1 the natural log of 1 of 0 so we have a 0 here so that is a 0 minus 1 minus 0 so that's going to be another 0 minus 0 so we get a numerator and we get a 0 in the numerator and in the denominator we get a 1 minus 1 which is 0 times the natural log of 1 which is 0 so 0 times 0 that is 0 and there you have it we have the indeterminate form that we need for law Patel's rule assuming that if we take the derivative of that and we put it over the derivative of that that that limit exists so let's try to do it so this is going to be equal to if the limit exists this is going to be equal to the limit as X approaches 1 and let's take the derivative in magenta I'll take the derivative of this numerator right over here and for this first term just do the product rule derivative of X is 1 and then so 1 times the natural log of X the derivative of the first term times the second term and then we're going to have plus the derivative of the second term plus 1 over x times the first term it's just the product rule so 1 over x times X we're going to see that's just 1 and then we have minus minus the derivative of X minus 1 well the derivative of X minus 1 is just 1 so it's just going to be minus 1 and then all of that is over all of that is over the derivative of this thing so let's take the derivative of that over here so the derivative of the first term of X minus 1 is just 1 multiply that times the second term you get natural log of X and then plus the derivative of the second term derivative of natural log of X is 1 over X times X minus one x times X minus one I think we can simplify this a little bit this one over x times X that's a 1 we're going to subtract 1 from it so these cancel out right there and so this whole expression can be rewritten as the limit as X approaches 1 the numerator is just natural log of X do that in magenta natural log of X and the denominator is the natural log of X plus X minus 1 over X so let's try to evaluate this limit here so if we take X approaches 1 of natural log of X that will give us a 1 natural log of 0 of 1 is 0 and over here we get natural log of 1 which is natural log of 1 is 0 and then plus 1 minus 1 over plus 1 minus 1 over 1 well that's just going to be another 0 or 1 minus 1 is 0 so you go to 0 plus 0 so you're gonna go to 0 over 0 again 0 over 0 so once again let's apply l'hopital's rule again let's take the derivative of that put it over the derivative of that so this if we're ever going to get to a limit is going to be equal to the limit the limit as X approaches 1 of the derivative of the numerator 1 over X right the derivative of Ln of X is 1 over x over the derivative of the denominator and what's that well derivative of natural log of X is 1 over X plus derivative of X minus 1 over X all right active you could view it this way as 1 over X times X minus 1 well derivative of X to the negative 1 will take the derivative of the first 1 times the second thing and then the derivative of the second thing times the first thing so the derivative of this first term X to the negative 1 is negative x to the negative 2 times the second term times X minus 1 plus the derivative of the second term which is just 1 times the first term plus one over X so this is going to be equal to I just had a random thing pop off into my computer sorry for that little sound if you heard it but where was I oh let's just simplify this over here we were doing our lop italo's rule so this is going to be equal to this is going to be equal to let me this is going to be equal to if we evaluate it X is equal to 1 the numerator is just 1 over 1 which is just 1 so we're definitely not going to have it indeterminate or at least a 0 over 0 form anymore and the denominator is going to be if you evaluate 1 its this is 1 over 1 which is 1 plus negative 1 to the negative 2 so or you say 1 to the negative 2 is just 1 and it's just a negative 1 but then you multiply that times 1 minus 1 which is 0 so this whole term is going to cancel out and then you have a plus another 1 over 1 so plus 1 and so this is going to be equal to 1/2 and there you have it using l'hopital's rule on a couple of steps we solve something that least initially didn't look like it was 0 over 0 we just added the two terms got 0 over 0 took derivatives of the numerators and denominators 2 times in a row to eventually get our limit