L'Hôpital's Rule
L'Hôpital's Rule Example 2 L'Hôpital's Rule Example 2
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- We need to evaluate the limit, as x approaches infinity, of 4x
- squared minus 5x, all of that over 1 minus 3x squared.
- So infinity is kind of a strange number.
- You can't just plug in infinity and see what happens.
- But if you wanted to evaluate this limit, what you might try
- to do is just evaluate-- if you want to find the limit as this
- numerator approaches infinity, you put in really large numbers
- there, and you're going to see that it approaches infinity.
- That the numerator approaches infinity as
- x approaches infinity.
- And if you put really large numbers in the denominator,
- you're going to see that that also-- well,
- not quite infinity.
- 3x squared will approach infinity, but we're
- subtracting it.
- If you subtract infinity from some non-infinite number, it's
- going to be negative infinity.
- So if you were to just kind of evaluate it at infinity,
- the numerator, you would get positive infinity.
- The denominator, you would get negative infinity.
- So I'll write it like this.
- Negative infinity.
- And that's one of the indeterminate forms
- that L'Hopital's Rule can be applied to.
- And you're probably saying, hey, Sal, why are we even
- using L'Hopital's Rule?
- I know how to do this without L'Hopital's Rule.
- And you probably do, or you should.
- And we'll do that in a second.
- But I just wanted to show you that L'Hopital's Rule also
- works for this type of problem, and I really just wanted to
- show you an example that had a infinity over negative
- or positive infinity indeterminate form.
- But let's apply L'Hopital's Rule here.
- So if this limit exists, or if the limit of their derivatives
- exist, then this limit's going to be equal to the limit as x
- approaches infinity of the derivative of the numerator.
- So the derivative of the numerator is-- the derivative
- of 4x squared is 8x minus 5 over-- the derivative of the
- denominator is, well, derivative of 1 is 0.
- Derivative of negative 3x squared is negative 6x.
- And once again, when you evaluated infinity, the
- numerator is going to approach infinity.
- And the denominator is approaching negative infinity.
- Negative 6 times infinity is negative infinity.
- So this is negative infinity.
- So let's apply L'Hopital's Rule again.
- So if the limit of these guys' derivatives exist-- or the
- rational function of the derivative of this guy divided
- by the derivative of that guy-- if that exists, then this
- limit's going to be equal to the limit as x approaches
- infinity of-- arbitrarily switch colors-- derivative
- of 8x minus 5 is just 8.
- Derivative of negative 6x is negative 6.
- And this is just going to be-- this is just a constant here.
- So it doesn't matter what limit you're approaching, this is
- just going to equal this value.
- Which is what?
- If we put it in lowest common form, or simplified
- form, it's negative 4/3.
- So this limit exists.
- This was an indeterminate form.
- And the limit of this function's derivative over this
- function's derivative exists, so this limit must also
- equal negative 4/3.
- And by that same argument, that limit also must be
- equal to negative 4/3.
- And for those of you who say, hey, we already
- knew how to do this.
- We could just factor out an x squared.
- You are absolutely right.
- And I'll show you that right here.
- Just to show you that it's not the only-- you know,
- L'Hopital's Rule is not the only game in town.
- And frankly, for this type of problem, my first reaction
- probably wouldn't have been to use L'Hopital's Rule first.
- You could have said that that first limit-- so the limit as x
- approaches infinity of 4x squared minus 5x over 1 minus
- 3x squared is equal to the limit as x approaches infinity.
- Let me draw a little line here, to show you that this is equal
- to that, not to this thing over here.
- This is equal to the limit as x approaches infinity.
- Let's factor out an x squared out of the numerator
- and the denominator.
- So you have an x squared times 4 minus 5 over x.
- Right? x squared times 5 over x is going to be 5x.
- Divided by-- let's factor out an x out of the numerator.
- So x squared times 1 over x squared minus 3.
- And then these x squareds cancel out.
- So this is going to be equal to the limit as x approaches
- infinity of 4 minus 5 over x over 1 over x squared minus 3.
- And what's this going to be equal to?
- Well, as x approaches infinity-- 5 divided by
- infinity-- this term is going to be 0.
- Super duper infinitely large denominator,
- this is going to be 0.
- That is going to approach 0.
- And same argument.
- This right here is going to approach 0.
- All you're left with is a 4 and a negative 3.
- So this is going to be equal to negative, or 4 over a
- negative 3, or negative 4/3.
- So you didn't have to do use L'Hopital's Rule
- for this problem.
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