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:
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 thethelimit 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+2−622−4 = 00
Which is indeterminate, so we are stuck. Or are we?
Let"s try L"Hôpital!
Differentiate both top và bottom (see Derivative Rules):
limx→2x2+x−6x2−4 = limx→22x+1−02x−0
Now we just substitute x=2 to lớn get our answer:
limx→22x+1−02x−0 = 54
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:
limx→∞exx2 = ∞∞
Both head lớn infinity. Which is indeterminate.
But let"s differentiate both top và bottom (note that the derivative of ex is ex):
limx→∞exx2 = limx→∞ex2x
Hmmm, still not solved, both tending towards infinity. But we can use it again:
limx→∞exx2 = limx→∞ex2x = limx→∞ex2
Now we have:
limx→∞ex2 = ∞
It has shown us that ex grows much faster than x2.
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:
00 ∞∞ 0×∞ 1∞ 00 ∞0 ∞−∞
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 ExistThis limit must exist:
Why? Well a good example is functions that never settle lớn a value.
Which is a ∞∞ case. Let"s differentiate top & bottom:
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.
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BUT we can bởi this:
limx→∞x+cos(x)x = limx→∞(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.