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We want to figure out the limit as x approaches 1 of the

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expression x over x minus 1 minus 1 over the

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natural log of x.

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So let's just see what happens when we just

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try to plug in the 1.

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What happens if we evaluate this expression at 1?

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Well then, we're going to get a one here, over 1 minus 1.

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So we're going to get something like a 1 over a 0, minus 1

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over, and what's the natural log of 1?

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e to the what power is equal to one?

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Well, anything to the zeroth power is equal to 1, so e to

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the zeroth power is going to be equal to 1, so the natural

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log of 1 is going to be 0.

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So we get the strange, undefined 1 over

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0 minus 1 over 0.

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It's this bizarre-looking undefined form.

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But it's not the indeterminate type of form that we looked

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for in l'Hopital's rule.

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We're not getting a 0 over a 0, we're not getting an

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infinity over an infinity.

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So you might just say, hey, OK, this is a non-l'Hopital's

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rule problem.

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We're going to have to figure out this limit some other way.

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And I would say, well don't give up just yet!

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Maybe we can manipulate this algebraically somehow so that

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it will give us the l'Hopital indeterminate form, and then

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we can just apply the rule.

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And to do that, let's just see, what happens if we

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add these two expressions?

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So if we add them, so this expression, if we add it, it

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will be, well, the common denominator is going to be x

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minus 1 times the natural log of x.

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I just multiplied the denominators.

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And then the numerator is going to be, well, if I multiply

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essentially this whole term by natural log of x, so it's going

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to be x natural log of x, and then this whole term I'm going

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to multiply by x minus one.

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So minus x minus 1.

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And you could break it apart and see that this expression

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and this expression are the same thing.

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This right here, that right there, is the same thing as x

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over x minus 1, because the natural log of x's cancel out.

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Let me get rid of that.

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And then this right here is the same thing as 1 over natural

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log of x, because the x minus 1's cancel out.

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So hopefully you realize, all I did is I added

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these two expressions.

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So given that, let's see what happens if I take the limit as

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x approaches 1 of this thing.

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Because these are the same thing.

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Do we get anything more interesting?

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So what do we have here?

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We have one times the natural log of 1.

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The natural log of 1 is 0, so we have 0 here, so that is a 0.

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Minus 1 minus 0, so that's going to be another 0, minus 0.

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So we get a 0 in the numerator.

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And in the denominator we get a 1 minus 1, which is 0, times

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the natural log of 1, which is 0, so 0 times 0, that is 0.

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And there you have it.

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We have indeterminate form that we need for l'Hopital's rule,

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assuming that if we take the derivative of that, and put it

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over the derivative of that, that that limit exists.

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So let's try to do it.

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So this is going to be equal to, if the limit exists, this

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is going to be equal to the limit as x approaches 1.

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And let's take the derivative in magenta, I'll take

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the derivative of this numerator right over here.

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And for this first term, just do the product rule.

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Derivative of x is one, and then so 1 times the natural log

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of x, the derivative of the first term times

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the second term.

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And then we're going to have plus the derivative of the

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second term plus 1 over x times the first term.

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It's just the product rule.

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So 1 over x times x, we're going to see, that's just 1,

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and then we have minus the derivative of x minus 1.

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Well, the derivative of x minus 1 is just 1, so it's just

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going to be minus 1.

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And then, all of that is over the derivative of this thing.

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So let's take the derivative of that, over here.

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So the derivative of the first term, of x minus 1, is just 1.

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Multiply that times the second term, you get natural log of x.

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And then plus the derivative of the second term, derivative

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of natural log of x is one over x, times x minus 1.

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I think we can simplify this a little bit.

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This 1 over x times x, that's a 1.

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We're going to subtract one from it.

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So these cancel out, right there.

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And so this whole expression can be rewritten as the limit

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as approaches 1, the numerator is just natural log of x, do

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that in magenta, and the denominator is the natural log

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of x plus x minus 1 over x

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So let's try to evaluate this limit here.

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So if we take x approaches one of natural log of x, that

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will give us a, well, natural log of 1 is 0.

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And over here, we get natural log of 1, which is 0.

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And then plus 1 minus 1 over plus 1 minus 1 over 1, well,

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that's just going to be another 0.

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1 minus 1 is zero.

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So you're going to have 0 plus 0.

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So you're going to get a 0 over 0 again.

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0 over 0.

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So once again, let's apply l'Hopital's rule again.

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Let's take the derivative of that, put it over

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the derivative of that.

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So this, if we're ever going to get to a limit, is going to be

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equal to the limit as x approaches 1 of the derivative

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of the numerator, 1 over x, right, the derivative of ln of

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x is 1/x, over the derivative of the denominator.

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And what's that?

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Well, derivative of natural log of x is 1 over x plus

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derivative of x minus 1 over x.

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You could view it this way, as 1 over x times x minus 1.

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Well, derivative of x to the negative 1, we'll take the

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derivative of the first one times the second thing, and

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then the derivative of the second thing times

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the first thing.

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So the derivative of the first term, x to the negative 1, is

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negative x to the negative 2 times the second term, times x

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minus 1, plus the derivative of the second term, which is

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just 1 times the first term, plus 1 over x.

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So this is going to be equal to, I just had a random thing

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pop up on my computer.

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Sorry for that little sound, if you heard it.

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But where was I?

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Oh, let's just simplify this over here.

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We were doing our l'Hopital's rule.

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So this is going to be equal to, let me, this is going to be

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equal to, if we evaluate x as equal to 1, the numerator is

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just 1/1, which is just 1.

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So we're definitely not going to have an indeterminate or

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at least a 0/0 form anymore.

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And the denominator is going to be, if you evaluate it at 1,

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this is 1/1, which is 1, plus negative 1 to the negative 2.

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So, or you say, 1 to the negative 2 is just 1, it's

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just a negative one.

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But then you multiply that times 1 minus 1, which is

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0, so this whole term's going to cancel out.

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And you have a plus another 1 over 1.

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So plus 1 And so this is going to be equal to 1/2.

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And there you have it.

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Using L'Hopital's rule and a couple of steps, we solved

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something that at least initially didn't look

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like it was 0/0.

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We just added the 2 terms, got 0/0, took derivatives of the

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numerators and the denominators 2 times in a row to

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eventually get our limit.

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