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Remember that the value of a definite integral is, by definition, a limit of a fairly complicated type, but this is not the kind of limit you will explore in this question. Instead, we will actually take limits of definite integrals, in the following sense. If the integrand or the limits in a definite integral depend on a parameter, then we can take the limit of the definite integral as the parameter tends to some fixed value or, in our case, tends to infinity. The examples in this Question of the Month were chosen to emphasize that you need to be careful when you take the limit of an integral that depends on a parameter. You may find that your intuition is not always reliable. On another level, the parts of this problem will give students a chance to think about and work with limits and definite integrals in new contexts, an experience that may lead to deeper mastery of both topics.
Before presenting our first example, we need to verify an important and useful fact, one that you may already know. We need to know that 
There are many ways to show this. Perhaps the easiest, but definitely not the most elementary, is to use l'Hôpital's Rule in the following indirect fashion. For 
let 
, so that 
We note that as n goes to infinity, both the numerator and the denominator of this fraction go to infinity, so l'Hôpital's Rule does apply. We actually need to apply l'Hôpital's Rule twice, since after one differentiation (with respect to n, of course), we still have an indeterminate form. After the second differentiation, we have
, and since y > 1, this fraction clearly goes to 0 as n goes to infinity, and the argument is complete.
As it happens, we have proved more than claimed. In fact, it follows that if
, then since 
It is also worth noting that the argument used above can be extended to show that for any value of p,
We will not need this, however. The only values of p we care about are 1/2, 1, and 3/2.
The first definite integral we consider is
(n is the parameter). Note that in light of the discussion above, the integrand goes to 0 for each x in the interval other than x = 1, in which case the limit of the integrand is 1. Based on this observation, we would expect that the limit of the integral would be 0 as n goes to infinity. Your first problem is to verify this fact.
(A) Show that 
The value of this limit agreed with our intuition. That may not be the case in the next three examples, which constitute the second part of this question. In these examples, the integrand has an infinite limit at one point in the interval [0,1], namely at 1, even though, using the result above, we see that the integrand has limit 0 at every other point in the interval.
(B) Find the limits of the integrals 
You should have discovered that the three limits are different, even though the limiting values of the integrand are identical. This shows that things can get complicated when we take the limit of a definite integral.
We move on to a more complex situation -- now both the integrand and one of the limits of the definite integral depend on the parameter. We note first, however, that
(*)
This is easy to see, using l'Hôpital's Rule, much as we did above, to find the limit of the fraction. One might say that the limit of the integral is infinite because the values of the integrand are too big on too long an interval. Contrast the behavior of the integral in (*) with the limit in the following problem, which involves a more complicated limit than the limits you found in (A) and (B).
(C) Show that the limit of the integral 
is 0.
In view of what you found in part (A) and the fact that the upper limit in the integral in part (C) tends to 1, the result in part (C) is not surprising. However, there is a dramatic contrast between the result in part (C) and what you will discover when you do the following, final part of this problem. Note first that
and hence
so that the upper limits of both of the integrals in part (D) tend to 1. (The second limit follows from the first since the second expression is just the square of the first. To establish the first limit, apply l'Hôpital's Rule to the logarithm of the expression, i.e., to
You should get 0, and if the logarithm of a quantity goes to 0, the quantity itself goes to 1. Note also that
if n > 1 and that
.)
(D) Find the limits of the integrals 
You should have discovered that the two integrals have different limits, each different from the limit in part (C), illustrating that even small changes in the limits of integration can have big effects on the limit of the integral.
The issues raised by the examples in this Question of the Month are addressed in a systematic way in a course in real analysis, which is typically offered to junior and senior mathematics majors. The kinds of examples we have seen here come up in the unit on sequences of functions.
Click here to view the answers!
Ben Klein is currently the Beverly F. Dolan Professor of Mathematics at Davidson College in Davidson, North Carolina, where he has taught since 1971. Ben's relationship with AP Calculus began in 1990 when he served as a Reader at Clemson University. He has attended every Reading since then and has served as a Table Leader in recent years. He just completed a four-year term on the AP Calculus Development Committee.
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