L'Hôpital's Rule
L'Hôpital's Rule Example 3 L'Hôpital's Rule Example 3
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- We want to figure out 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 a one 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 one?
- Well, anything to the zeroth power is equal to 1, so e to
- the zeroth 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 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 l'Hopital's rule.
- We're not getting a 0 over a 0, we're not getting an
- infinity over an infinity.
- So you might just say, hey, OK, this is a non-l'Hopital's
- rule problem.
- We're going to have to figure out this limit some other way.
- And I would say, well don't give up just yet!
- Maybe we can manipulate this algebraically somehow so that
- it will give us the l'Hopital indeterminate form, and then
- we can just apply the rule.
- And to do that, let's just see, what happens if we
- add these two expressions?
- So if we add them, so this expression, if we add it, it
- will be, well, the common denominator is going to be x
- minus 1 times the natural log of x.
- I just multiplied the denominators.
- And then the numerator is going to be, well, if I multiply
- essentially this whole term by natural log of x, so it's going
- to be x natural log of x, and then this whole term I'm going
- to multiply by x minus one.
- So minus x minus 1.
- 41 00:01:58,51 --> 00:02:00,54 And you could break it apart and see that this expression
- and this expression are the same thing.
- Let me get rid of that.
- And then this right here is the same thing as 1 over natural
- log of x, because the x minus 1's 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 one times the natural log of 1.
- The natural log of 1 is 0, so we have 0 here, so that is a 0.
- Minus 1 minus 0, so that's going to be another 0, minus 0.
- the natural log of 1, which is 0, so 0 times 0, that is 0.
- And there you have it.
- We have indeterminate form that we need for l'Hopital's rule,
- assuming that if we take the derivative of that, and 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 one, 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 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 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 one over x, times x minus 1.
- 88 00:04:32,14 --> 00:04:34,24 I think we can simplify this a little bit.
- This 1 over x times x, that's a 1.
- We're going to subtract one from it.
- So these cancel out, right there.
- And so this whole expression can be rewritten as the limit
- as approaches 1, the numerator is just natural log of x, do
- that in magenta, 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 one of natural log of x, that
- will give us a, well, natural log of 1 is 0.
- And over here, we get natural log of 1, which is 0.
- And then plus 1 minus 1 over plus 1 minus 1 over 1, well,
- that's just going to be another 0.
- 1 minus 1 is zero.
- So you're going to have 0 plus 0.
- So you're going to get a 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 as x approaches 1 of the derivative
- of the numerator, 1 over x, right, the derivative of ln of
- x is 1/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.
- You could view it this way, as 1 over x times x minus 1.
- Well, derivative of x to the negative 1, we'll take the
- derivative of the first one times the second thing, and
- then the derivative of the second thing times
- the first thing.
- So the derivative of the 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 1 over x.
- So this is going to be equal to, I just had a random thing
- pop up on 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 l'Hopital's rule.
- So this is going to be equal to, let me, this is going to be
- equal to, if we evaluate x as equal to 1, the numerator is
- just 1/1, which is just 1.
- So we're definitely not going to have an indeterminate or
- at least a 0/0 form anymore.
- And the denominator is going to be, if you evaluate it at 1,
- this is 1/1, which is 1, plus negative 1 to the negative 2.
- So, or you say, 1 to the negative 2 is just 1, it's
- just a negative one.
- But then you multiply that times 1 minus 1, which is
- 0, so this whole term's going to cancel out.
- And 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 and a couple of steps, we solved
- something that at least initially didn't look
- like it was 0/0.
- We just added the 2 terms, got 0/0, took derivatives of the
- numerators and the denominators 2 times in a row to
- eventually get our limit.
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