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

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In symbols we can write:

*lim***x→c***f(x)***g(x)** = *lim***x→c***f’(x)***g’(x)**

The limit as x approaches c of "f-of−x over g-of−x" equals the**thelimit 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 lớn take the derivative.**

**At x=2** we would normally get:

*22+2−6***22−4** = *0***0**

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

Let"s try L"Hôpital!

Differentiate both top và bottom (see Derivative Rules):

*lim***x→2***x2+x−6***x2−4** = *lim***x→2***2x+1−0***2x−0**

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

*lim***x→2***2x+1−0***2x−0** = *5***4**

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

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

Normally this is the result:

*lim***x→∞***ex***x2** = *∞***∞**

Both head lớn infinity. Which is indeterminate.

But let"s differentiate both top và bottom (note that the derivative of ex is ex):

*lim***x→∞***ex***x2** = *lim***x→∞***ex***2x**

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

*lim***x→∞***ex***x2** = *lim***x→∞***ex***2x** = *lim***x→∞***ex***2**

Now we have:

*lim***x→∞***ex***2** = ∞

It has shown us that ex grows much faster than x2.

## Cases

We have already seen a *0***0** and *∞***∞** example. Here are all the indeterminate forms that L"Hopital"s Rule may be able to help with:

*0***0** *∞***∞** 0×∞ 1∞ 00 ∞0 ∞−∞

## 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) is not equal to zero either side of c.

### The Limit Must Exist

This limit must exist:*lim***x→c***f’(x)***g’(x)**

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

Which is a *∞***∞** case. Let"s differentiate top & bottom:

*lim***x→∞***1−sin(x)***1**

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

So that new limit does not exist!

**And so L"Hôpital"s Rule is not usable in this case.Xem thêm: Army Phiên Bản Căn Góc V9 Chuẩn Full Nhân Vật, Army 239 V11 Căn Góc 360 Độ**

BUT we can bởi this:

*lim***x→∞***x+cos(x)***x** = *lim***x→∞**(1 + *cos(x)***x**)

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