Warning:(.dvi version)(.pdf version) <--(READ ME)
When NOT to use l'Hôpital's ruleThe most important thing to learn about l'Hôpital's rule is when it should not be used:
- When the limits of the two parts are not both 0, or both infinite. In this case the rule is likely to give a wrong answer! Example:
limx->0+ (cos x)/x
is positive infinity, because the numerator approaches 1 while the denominator approaches 0. If we incorrectly apply l'Hôpital's rule, we get
limx->0+ (- sin x)/1 = 0.
- When it is making the problem worse. Example:(.dvi version)(.pdf version) <--(READ ME)
- When there is a better way to get the answer.
Example 1: The limit at infinity of
(x2 - 1)/(2x2 + 1)
can be correctly evaluated by two successive applications of l'Hôpital's rule; but it can also be found by a rule we learned earlier: Divide both numerator and denominator by the highest power (x2) and take the limit directly, getting 1/2 very quickly.
Example 2: Consider the limit at 0 of
(cos x - 1)/x.
Since top and bottom both approach 0, it is permissible to use the rule and get
limx->0 (- sin x)/1 = 0.
However, it is less mysterious and more instructive to recall (from the section on differentials) that the cosine function has the quadratic approximation
cos x ÷ 1 - x2/2
and therefore the numerator behaves near 0 like - x2/2. This obviously vanishes faster than the denominator, x, so the limit is zero.
Clearly, the first two items on this list are absolute prohibitions, while the last one is merely friendly advice. Many students overuse l'Hôpital's rule, relying on it as a "black box" when they would learn much more (and solve the problems equally fast) by just taking a close look at, and comparing, the behavior of the numerator and denominator as x -> a. In particular, you should first ask yourself whether you know the linear or quadratic approximation of the numerator or denominator around x = a. If so, then it will probably become clear that the fraction is approximately equal to a constant times some power of (x - a), and the limit is infinite, finite, or zero depending on whether the exponent is negative, zero, or positive (see Example 2 above). Similarly, when a limit is taken at infinity, ask whether the numerator or denominator "behaves like" a certain power of x as x becomes large; the ratio of two powers approaches an obvious limit at infinity, depending on the two exponents (as in Example 1).
When SHOULD you use l'Hôpital's rule?By far the most important situation for using the rule is when the numerator or denominator does not have an obvious power-like behavior as x approaches a. This is the case for the logarithm function as x approaches either 0 or infinity, and for the exponential function as x approaches either positive or negative infinity. (That is why the section on l'Hôpital's rule is in this chapter!)
As homework you have a number of limits of ratios of exponentials, logarithms, and ordinary powers, which you should evaluate for practice. After awhile, though, the results of such calculations become very predictable. They can be summarized in a list of general conclusions:
- As x approaches positive infinity, ex increases faster than any power, xn.
- As x approaches positive infinity, e-x decreases faster than any negative power, x-n.
- As x approaches positive infinity, ln x, although it goes to infinity, increases more slowly than any positive power, xa (even a fractional power such as a = 1/200).
- As x -> 0+, - ln x goes to infinity, but more slowly than any negative power, x-a (even a fractional one).
The exponential indeterminate forms, and the 00 controversy
The limit (as x approaches a) of f(x)g(x) can't always be found just by taking the limits of f and g individually -- just as the limit of f(x)/g(x) is not the quotient of the individual limits if those limits are both 0 or infinity. To see when and why there is a problem, note that the logarithm of f(x)g(x) is g(x) ln f(x), and consider the limit of that. If g approaches 0 and the logarithm approaches positive or negative infinity, or if g approaches infinity and the logarithm approaches 0, then we have an indeterminate form of the type "0 times infinity". (What do we do then?) These three cases correspond, back in the original function, to
- g -> 0, f -> infinity;
- g -> 0, f -> 0;
- g -> infinity, f -> 1.
The rest of this page is just an interesting side remark, not important to the course.
You may be surprised to see the second item in the list, since 00 at first sight looks like a fairly tame object. In fact, however, if you take a look on any given day at an Internet newsgroup devoted to discussion of mathematics, you are likely to find a lively debate going on about whether 00 is equal to 1, or is undefined. (Sometimes these arguments are started by a "crank" who charges that the mathematics profession is covering up some embarrassing scandal associated with the meaning of 00.)
To see why there is a problem, note first that
- x0 = 1 for all nonzero x;
- 0y = 0 for all positive y;
- 0y is undefined (or infinite) for negative y.
Argument in favor of defining 00 to be 1: There are many useful formulas involving expressions of the form xy that remain meaningful and true when x and y equal 0 if 00 is intepreted as 1. The simplest example is the binomial formula
(a + b)2 = a2b0 + 2a1b1 + a0b2.
A related example is Taylor's theorem, which involves powers of (x - a).
Argument in favor of leaving 00 undefined: If f(a) = 0 = g(a), and both functions are continuous, then we would be tempted to jump to the conclusion that
limx -> a f(x)g(x) = 1,
forgetting that 00 really is an indeterminate form. To prevent confusion, we outlaw 00.
More information about this topic from the sci.math Frequently Asked Questions list
Something to think about: Why don't we define 0/0 to be 1, instead of insisting that 0/0 is undefined? (an answer)
Офицер гордо кивнул: - Да. Когда церковь получит все останки этого великого человека, она причислит его к лику святых и разместит отдельные части его тела в разных соборах, чтобы все могли проникнуться их величием. - А у вас здесь… - Беккер не сдержал смешка.