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L'Hôpital's rule
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Evaluating indeterminate limits of the form x→alimg(x)f(x) when both numerator and denominator approach 0 or ∞, by replacing them with x→alimg′(x)f′(x) and repeating if necessary.
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Caroline and Carmen are racing cars on a track. Caroline begins 85km from the finish line, which she drives toward in a car at a constant speed 170km/h. Carmen has a racecar, which travels at a constant speed of 340km/h,and she starts 170km from the finish line.
Notice that Carmen is always twice as far from the finish line as Caroline, even though they both reach it at the same time. This illustrates a very powerful:
If two functions f and g both approach zero as x approaches a certain value, then the ratio of the two functions approaches the ratio of their derivatives.
This works even when the speed is not constant, because very close to the limit, speed becomes essentially constant:
This also works when f and g both tend to infinity: the long term behavior of the derivative determines the long term behavior of the function:
More concretely, the ratio of redblue changes erratically for small values (x<5), but we clearly see that the ratio stabilizes for x>20.
L'Hôpital's rule
AHL 5.12
L'Hopital's rule states that for a limit of the form x→alimg(x)f(x), if both f(x) and g(x) approach zero or both approach infinity, then the value of the limit is the same as the value of x→alimg′(x)f′(x).
In intuitive terms, if the limit is indeterminate, we can take the derivative of the top and the bottom and evaluate that limit.
Sometimes, the process of taking derivatives will need to be repeated multiple times before the limit becomes determinate.
Example: x→0limx2ex−1−x
First, check if the limit is indeterminate:
e0−1−0=0
and
02=0
So it is inderterminate. We can apply L'Hopital's rule:
x→0limx2ex−1−x=x→0lim2xex−1
Which is still indeterminate since e0−1=2⋅0=0. Applying L'Hopital's rule again:
x→0limx2ex−1−x=x→0lim2ex
This is no longer indeterminate:
x→0lim2ex=2e0=21
Note, sometimes it's easier to factor out common terms than to use l'Hôpital's rule:
3x+13x2+x=3x+1x(3x+1)=x,
so
x→−31lim3x+13x2+x=−31.
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