Hello and welcome to this video about L’Hôpital’s Rule!
When taking certain types of limits, you’ll find this 300-year-old rule can come in extremely handy.
Guillaume François Antoine de l'Hôpital was a French mathematician in the late 1600s who rubbed elbows with the likes of the Bernoulli brothers
and one of the fathers of calculus himself, Gottfried Leibniz.
In 1696, when he was about 35 years old, he published his book Infinitesimal Calculus with Applications to Curved Lines, the first textbook on infinitesimal calculus,
and the first book to contain the famous rule.
Though the rule was given to him by Johann Bernoulli two years prior, stirring some controversy, it bears L’Hôpital’s name and is still in use today.
Let’s start by taking some limits and establishing where L’Hôpital’s Rule into play and why it is so important.
First, consider this expression:
the limit as x approaches 5, of x squared minus 25, over x minus 5.
Substituting 5 into the expression yields the indeterminate form 0 over 0,
but factoring the numerator results in (x plus 5) times (x minus 5) over (x minus 5),
which reduces to x plus 5.
Substitute 5 again to find a limit of 10.
Recall that an indeterminate form means that we cannot yet tell if a limit exists.
The idea is that if we substitute the value the input approaches and the result is indeterminate, do some algebra and substitute again.
If we get a limit (or there for sure isn’t one), great.
If the result is indeterminate, do some more algebra and try again.
Repeat as needed.
Take a look at this limit:
the limit as x approaches infinity, of 5x minus 2 x cubed, over 3 x cubed plus 1.
As x approaches infinity, the function approaches infinity over infinity,
indeterminate once again.
We are only concerned with the cubed terms in the numerator and denominator because they’ll grow the fastest and determine the overall limit.
So, only considering those, the limit becomes
the limit as x approaches infinity, of negative 2 x cubed, over 3 x cubed,
which is negative 2 over 3.
Unfortunately, these methods of finding limits do not always work.
Suppose we were going to evaluate this limit:
the limit
as x approaches 0,
of sin of x,
over x.
Substituting 0, we'll get 0 over 0.
No problem, right?
We got an indeterminate form, so all we need to do is a little bit of algebra.
But now we have a serious problem.
That quotient won’t simplify algebraically.
So we can’t yet tell if there’s a limit, but we have no algebraic way of finding one.
Or do we?
It’s easy to see on the graph that this limit is 1
(note that the function is undefined at x equals 0),
but of course graphs are not always practical.
L'Hopital's Rule says that we can do this:
We can take the limit as x approaches 0,
of the derivative of the numerator,
sine of x,
over the derivative of the denominator,
which is x.
Then we can simplify this.
So, the derivative of, we're going to keep the limit as x approaches 0, of the derivative of sine of x, is cosine of x.
And the derivative of x is 1.
So, if we plug in 0 for x, we get cosine of 0,
which is equal to 1.
It’s not a coincidence that we find a limit of 1, just as the graph showed.
The idea is not exactly intuitive at first, but with a 0 over 0 limit, it’s an application of local linearity.
Really, really close to x equals 0, sine x looks like the line tangent to sine x there.
In other words, since the derivative of sine x is cosine x, the graph of sine x close to 0 looks like its tangent line at 0, which has a slope of cosine of 0, or 1.
Of course, the function in the denominator, x, looks exactly the same.
Because of local linearity, the ratio of function derivatives is usually a very good approximation of the ratio of function values.
Take a look at this one.
When we go to take the limit of this expression,
the limit
as x approaches infinity
of e to the x
over 2 x squared,
we end up with the indeterminate form infinity over infinity with no way to simplify algebraically.
In the case of 0 over 0, we “zoom in” really close to where we need to look.
In an infinity over infinity case like this, we’ll use the same method, but look out at one of the ends.
Since both functions approach infinity as x approaches infinity, the question is: which one goes to infinity faster?
So, using L'Hopital's Rule, we can do this:
We have the limit as x approaches infinity
of the derivative of the numerator,
e to the x,
over the derivative of the denominator,
2 x squared,
which simplifies to, the limit as x approaches infinity,
of e to the x,
over 4x.
Since both functions still approach infinity, apply the rule again!
So we have the limit as x approaches infinity,
of the derivative of the numerator, e to the x,
over the derivative of the denominator, 4x,
which gives us, the limit
as x approaches infinity
of e to the x over 4.
Which, we know that this equals infinity
because e to the x is gonna grow way faster than 4.
After the second application, it’s clear that we have a limit of infinity.
Like many mathematical rules, L'Hopital's Rule won’t work 100% of the time.
But it is a powerful technique to learn because it will get you out of many tight spots!
Here is the rule formally stated:
If the limit as x approaches a, of f of x, over g of x, equals 0 over 0,
or the limit as x approaches a, of f of x, over g of x, equals positive or negative infinity over positive or negative infinity,
then the limit as x approaches a, of f of x, over g of x, is equal to the limit as x approaches a, of f prime of x, over g prime of x.
Let’s try a few more!
On the first attempt, the result is 0 over 0.
Time for L'Hopital's Rule!
So, we're gonna take the limit
as x approaches 2,
of the derivative of the numerator,
which is 3 x squared,
minus 14x, plus 10,
over the derivative of the denominator, which gives you 2x minus 1.
Then we're gonna plug in 2 anywhere there's an x, which will give us
negative 6 over 3,
which is equal to negative 2.
So our limit for this expression is negative 2.
Now let's try this expression.
So our first substitution yields 0 over 0.
Now we're going to use L'Hopital's Rule.
So, this is equal to the limit as x approaches negative 3,
of the derivative of the numerator,
which is pi, cosine of pi x,
over the derivative of the denominator, which is 2x.
Then, we'll plug in negative 3 anywhere we see an x, and that'll give us pi over 6.
So the limit of this expression is pi over 6.
Now let's try a natural log.
Both functions approach infinity as x approaches infinity.
But which one approaches infinity faster?
We can use L'Hopital's Rule to find out that the limit as t approaches infinity,
of the derivative of the numerator, 1 over t,
over the derivative of the denominator, 2t.
Which, then we can rearrange, to get the limit as t approaches infinity,
of 1 over 2 t squared.
And then we'll plug in infinity anywhere we see a t,
so, since the denominator is growing way faster than the numerator, we know that our limit is equal to 0.
So that’s L'Hopital's Rule, a valuable addition to any limit-taking repertoire.
I hope that this video helped you understand the rule, when to use it, and how it works!
Thanks for watching, and happy studying!