a) The problem was to use L'Hopital's Rule to determine the limit
ex - 1
lim
x > 0 x
Step 1: Determine that both numerator and denominator go to zero in
the limit. Since Step 2: Find the derivatives of the numerator and denominator.
The derivative of
Step 3: Apply L'Hopital's Rule. This means that
ex - 1 ex
lim = lim
x > 0 x x > 0 1
Step 4: Take the limit.
When x is zero, the right-hand expression above clearly goes
to
b) The problem was to find the limit
ex - 1 - x
lim
x > 0 x2
Step 1: Does it qualify for L'Hopital?
The numerator of this one differs from the numerator of the first one only
by x. And as x approaches zero, that's no difference at
all. So if
Step 2: Take the derivatives of numerator and denominator.
The derivative of
Step 3: Apply L'Hopital's Rule. That means that
ex - 1 - x ex - 1
lim = lim
x > 0 x2 x > 0 2x
Step 4: Take the limit. We ended up with a numerator and denominator that are again both zero in the limit. So we need to apply L'Hopital again. But observe that the expression to the right of the equal sign above is exactly one half of the expression you were asked to take the limit of in the first problem. And you know that that limit came out to be 1. So this limit, when you have applied L'Hopital for the second time, must come out to be 1/2.
c) The problem was to find the limit
ex - 1 - x - (1/2)x2
lim
x > 0 x3
The steps are the same as in the previous two problems. Observe that
when you take the derivatives of the numerator and denominator and
then apply L'Hopital's Rule, you get
ex - 1 - x - (1/2)x2 ex - 1 - x
lim = lim
x > 0 x3 x > 0 3x2
Observe that the expression to the right of the equal is exactly one third
of the expression you were asked to take the limit of in the last example.
The last example had a limit of 1/2 so this one must have a limit
of 1/6.
d) The problem was to find the limit
ex - 1 - x - (1/2)x2 - (1/6)x3
lim
x > 0 x4
Do you have any doubt that when you take the derivatives of numerator and
denominator and then apply L'Hopital's Rule that you will get anything
other than one fourth of the expression you were asked to find the limit
of in the last example? The last example had a limit of 1/6, so
this one must have a limit of 1/24.
So what is the pattern here? Clearly I am using the limit I get in each example to be the coefficient of the new term that I add to the numerator to get the next example. And in the denominator I am simply increasing the power of x by one with each successive example. So if there were an example e), it would be
ex - 1 - x - (1/2)x2 - (1/6)x3 - (1/24)x4
lim
x > 0 x5
And its limit would have to be one fifth of the limit we arrived at
for the last example (the one with the fourth power in the denominator).
Its limit was 1/24, so the limit here would be 1/120.
Recall that the limit in the first example was 1, which
is the same as 1/1!. The limit in
the second example was
Now think about what would have happened with that fifth example if you had started on it without ever seeing the other four. You would have had to apply L'Hopital's rule five times. So you would have had to take the derivative of x5 five times. And at the end of that you would have ended up with a constant of 5!. Here is an interesting and useful fact you can prove for yourself using induction: If you start with xn, where n is a whole number, and you take its derivative exactly n times, you will end up with a constant of n!.
See how that fact also plays into the choice of coefficients for
the numerator terms. In the nth example in this series
of examples, you will have to apply L'Hopital n times.
On the n-1st time, you will have to end up with a
numerator of
ex - 1 - x - (1/2)x2 - (1/6)x3 - ... - (1/(n-1)!)xn-1
lim
x > 0 xn
and its limit would be 1/n!, which would result from applying
L'Hopital n times.
We will be coming back to this line of thought in the next section. So go ahead and play with these examples until you the pattern here begins to seem intriguing.
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