So in this video. We're going to talk about L'Hopital's Rule and
L'hopital's rule is basically something to help calculate limits
and the idea is
You know x is approaching something it could be positive infinity. It could be negative infinity it could be some finite number
the idea is
Maybe the top is going to infinity positive or negative infinity [the] same thing with the denominator
So you end up with the form what's called the indeterminate form infinity over infinity
perhaps also the top is going to 0 and maybe the bottom is also going [to] 0
[it] says in that case which you can do is you can basically turn one limit problem into another limit problem by taking the derivative
of the top and dividing that by the derivative [of] the bottom
Ok and sometimes they'll put a little lh or h in between there to indicate that you're using L'hopital's rule
[be] careful to because you're taking the derivative of the numerator all by itself same thing with the denominator
you're not using the quotient rule when you do the derivative, so
Let's do some examples here
the limit as x goes to 0 of sine of x over
X you may remember this one
but if you plug in 0 notice, you get sine of 0 which is [0] you'll get [0] on the bottom which is 0
That means I can use L'hopital's rule on it
And I'll get the limit as x approaches 0 the derivative of sine is just cosine of x and
Then the derivative of x is 1 and we basically repeat this process we plug 0 in well cosine of 0 is 1 and
you're dividing by 1 and
That is your answer in this case
And you may recall that the limit as x [goes] to 0 of sine x over x is in fact equal to 1?
Let's do a few of these here
L'hopital's rule is one of these things [too] that makes a lot of limit problems easier
But don't get carried away you have to kind of pick and choose your battles people will try to use L'hopital's rule
[even] when it's not applicable
So really make sure that you're getting either infinity over infinity or zero over zero
let's make it the limit as x goes to infinity of Ln of x divided by x as
x goes to infinity
Ln of x is infinity same thing with the bottom so I'm getting them the form infinity over infinity
Which [means] I can use L'hopital's rule again
So I'll have the limit as x approaches infinity
the derivative of x is just 1 the derivative of Ln of x is 1 over x and
1 over x divided by 1 is the same thing [as] 1 divided by x
Well 1 over a large number is the same thing as having [0] so that'll be the [answer] to your limit
Let's do a slight variation on this problem. Let's make it the limit as x approaches 0 from the right of
Ln of X
over x
Ok so let's think about this now. Well certainly as x approaches 0 from the right the denominator is going to get close to 0
What happens to Ln of x though?
does that become 0 and again this is where it's maybe useful to think about the graph of Ln of x
Remember Ln of x looks like this
Roughly
So it says as x approaches 0 from the right Ln of x actually goes down to negative infinity
So you're kind of getting the form negative infinity over zero [which] is not one of the indeterminate forms so you can't use L'hopital's rule
The idea is though the numerator is getting big, but it's negative
The denominator is getting small
But it's positive well if you take a big number and divide it by a small number you still get a big number
The denominator again is positive the numerators numerator is negative
So your answer to this problem is actually just going to be negative infinity
[and] there's not a lot [of] work to do so again. Here's a little word of warning about
you know when to use L'hopital's Rule and when not
Otherwise you could end up getting something very incorrect out of there
Let's look maybe at one more problem here
Trying to find a more complicated one so how about the limit as x goes to infinity?
x divided by Ln of 1 plus
2 times E to the x
[okay], so as x goes to infinity the numerator is certainly going to Infinity E
To the x will go to the infinite it will go to infinity as x goes to [infinity]
so the inside of the natural logarithm is going to infinity and if you plug large numbers into the natural logarithm you get
Large numbers out so we are getting infinity over infinity
So I'm going to use L'hopital's rule here. [I'll] have the limit as again x goes to infinity the numerator is just 1
The denominator again if I take the derivative I'll get 1 over 1 plus 2 E to the x and
then I have to multiply that by the derivative of the inside which is just going to be 2 e to the x
And I'm going to clean this up [before] I do anything before I start thinking about my limit
So the denominator if you flip it and multiply, you'll have 1 plus 2 e to the x on
top and in the denominator, you'll have 2 e to the x
So again as x goes to infinity e to the x is infinity times 2 plus 1 you're still going to get infinity over infinity
Which means we can use L'hopital's rule again? So you can use L'hopital's rule as many times as you want
Well the derivative of 1 is 0 the derivative of 2 e to the x is just 2 e to the x?
Same thing on the bottom
[you] know again, you could think well. This is still infinity over infinity, and if I keep doing L'hopital's rule
[I'm] not going to get anywhere, but remember to simplify down at algebraically
- E to the [x] divided by 2 E to the [x] is just going to equal?
1 and the limit as x goes to infinity of 1 is simply 1 so that's going to be the answer in this case
Don't think I did anything
Weird on this problem [check] over everything
I think it looks okay to me here. So this is the basic idea if you get 0 [over] 0 or infinity over infinity
To start taking the derivative of the top the derivative of the bottom
simplify things down algebraically and
then just try to
Repeat the process if you're still getting 0 over 0 or infinity over infinity
Keep doing L'hopital's rule again until you basically don't get that form anymore