L'Hôpital's Rule

L'Hôpital's Rule can help us calculate a limit that may otherwise be hard or impossible.

L'Hôpital is pronounced "lopital". He was a French mathematician from the 1600s.

It says the limit when we divide one function by another is the same after we differentiate each function (with some special conditions shown later).

In symbols we can write:

limx→cf(x)g(x) = limx→cf'(x)g'(x)

The limit as x approaches c of "f-of−x over g-of−x" equals the
the limit as x approaches c of "f-dash-of−x over g-dash-of−x"

All we did is add that little dash mark ' on each function, which means to differentiate.

We differentiate the top and bottom separately. We do not use the Quotient Rule here.

Example:

limx→2x²+x−6x²−4

At x=2 we would normally get:

2²+2−62²−4 = 00

Which is indeterminate, so we are stuck. Or are we?

Let's try L'Hôpital!

Differentiate both top and bottom (see Derivative Rules):

limx→2x²+x−6x²−4 = limx→22x+1−02x−0

Now we just substitute x=2 to get our answer:

limx→22x+1−02x−0 = 54

Here's the graph, notice the "hole" at x=2:

Graph of (x^2+x-6) divided by (x^2-4) showing a hole at x=2 and a value of 1.25

Note: we can also get this answer by factoring, see Evaluating Limits.

Example:

limx→∞e^xx²

Normally this is the result:

limx→∞e^xx² = ∞∞

Both head to infinity. Which is indeterminate.

But let's differentiate both top and bottom (note that the derivative of e^x is e^x):

limx→∞e^xx² = limx→∞e^x2x

Hmmm, still not solved, both tending toward infinity. But we can use it again:

limx→∞e^xx² = limx→∞e^x2x = limx→∞e^x2

Now we have:

limx→∞e^x2 = ∞

It has shown us that e^x grows much faster than x².

Cases

We have already seen a 00 and ∞∞ example. Here are all the indeterminate forms that L'Hopital's Rule may be able to help with:

∞∞

0×∞

1^∞

0⁰

∞⁰

∞−∞

To use L'Hôpital's Rule on these other forms, we usually have to rearrange them into 00 or ∞∞ first.

Conditions

Differentiable

For a limit approaching c, the original functions must be differentiable either side of c, but not necessarily at c.

Likewise g'(x) isn't equal to zero either side of c.

The Limit Must Exist

This limit must exist:

limx→cf'(x)g'(x)

Why? Well a good example is functions that never settle to a value.

Example:

limx→∞x+cos(x)x

Which is a ∞∞ case. Let's differentiate top and bottom:

limx→∞1−sin(x)1

And because it just wiggles up and down it never approaches any value.

So that new limit doesn't exist!

And so L'Hôpital's Rule isn't usable in this case.

BUT we can do this:

limx→∞x+cos(x)x = limx→∞(1 + cos(x)x)

As x goes to infinity then cos(x)x tends to between −1∞ and +1∞, and both tend to zero.

And we are left with just the "1", so:

limx→∞x+cos(x)x = limx→∞(1 + cos(x)x) = 1

18014, 18015, 18016, 18017, 18018, 18019, 18020, 18021, 18023, 18024, 18025, 18026, 18027, 18029, 18030, 18032, 18033, 18022, 18028, 18031