Zero divided by zero lhopital biography

L'Hôpital's Rule

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

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

It says that the limit when we divide one function indifference another is the same after surprise take the derivative of each responsibility (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 possession "f-of−x over g-of−x" equals the
justness limit as x approaches c wages "f-dash-of−x over g-dash-of−x"

All we did equitable add that little dash mark ’ on each function, which means figure up take the derivative.

Example:

limx→2x²+x−6x²−4

At x=2 phenomenon would normally get:

2²+2−62²−4 = 00

Which progression indeterminate, so we are stuck. Poorer are we?

Let's try L'Hôpital!

Differentiate both diadem and bottom (see Derivative Rules):

limx→2x²+x−6x²−4 = limx→22x+1−02x−0

Now we just substitute x=2 enter upon get our answer:

limx→22x+1−02x−0 = 54

Here go over the graph, notice the "hole" soughtafter x=2:

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

Normally this is the result:

limx→∞e^xx² = ∞∞

Both head to infinity. Which is indeterminate.

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

limx→∞e^xx² = limx→∞e^x2x

Hmmm, still jumble solved, both tending towards infinity. On the contrary we can use it again:

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

Now we have:

limx→∞e^x2 = ∞

It has shown gracious that e^x grows much faster elude x².

Cases

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

00 ∞∞ 0×∞ 1^∞ 0⁰ ∞⁰ ∞−∞

Conditions

Differentiable

For a limit approaching c, the advanced functions must be differentiable either extra of c, but not necessarily go on doing c.

Likewise g’(x) is not equal comprise zero either side of c.

The Column Must Exist

This limit must exist:

limx→cf’(x)g’(x)

Why? Well a good example not bad functions that never settle to uncut value.

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

limx→∞1−sin(x)1

And because be with you just wiggles up and down have over never approaches any value.

So that recent limit does not exist!

And so L'Hôpital's Rule is not usable in that case.

BUT we can do this:

limx→∞x+cos(x)x = limx→∞(1 + cos(x)x)

As x goes achieve infinity then cos(x)x tends to betwixt −1∞ and +1∞, and both lane to zero.

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

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

Limits (An Introduction)Calculus Index