f(x) lim does not exist.

Transcription

1 Indeterminate Forms and L Hopital s Rule When we computed its of quotients, i.e. its of the form f() a g(), we came across several different things that could happen: f(). a g() = f(a) g(a) when g(). a f() 2. g() =, but f(), so a a a g() does not eist. f() 3. g() = = f(). Then a a a g() decide for certain. may or may not eist; we had to do more work to The form of the third it is called an indeterminate form; basically, we don t know what an answer of this form means. L Hopital s Rule, which we will learn in this section, will give us a method for handling its that result in indeterminate forms. Indeterminate Forms The following epressions, known as indeterminate forms, often come up when attempting to evaluate a it using the evaluation technique:, ± ±,, ±, ±,,. By indeterminate form, we mean a form that doesn t make sense; we don t know what means since c = if c and a is undefined if a. The following rule can help us evaluate its that result in such indeterminate forms: Theorem. (L Hopital s Rule) If the it a f() g() results in one of the indeterminate forms if the second it eists. f() a g() = f () a g (), or ± ±, then The it above can be a normal it, a one-sided it, or even a it as approaches or.

2 In other words, if we try evaluating a it by evaluation and end up with one of the indeterminate forms ± or ±, then we can differentiate both the numerator and denominator of the original fraction and try evaluating the it of the new fraction. Notice that L Hopital s Rule has NOTHING to do with the quotient rule. We use the quotient rule to find the derivative of a quotient of two functions. We use L Hopital s Rule to find the it of a quotient of two functions. Note also that L Hopital s Rule ONLY applies when the it in question gives us one of the two specified indeterminate forms. sin Eample. Find. Evaluation gives us sin =, so we try using L Hopital s Rule: sin LR cos = = cos =. (ln ) 2 Eample. Find. Again, evaluation gives us an indeterminate form, this time, so we apply L Hopital s Rule: (ln ) 2 2 ln = but we still have an indeterminate form so we apply L Hopital s Rule again! 2 2 ln LR = = 2 =. Indeterminate Products We cannot apply L Hopital s Rule directly to an indeterminate form like, but we can rewrite it so that L Hopital s Rule applies. In such a case, rewriting one of the factors as a fraction can be quite helpful. Eample. Let s calculate ln (which results in the form ). Before making the calculation, let s use a bit of intuition to guess at the answer. The function ln grows more slowly than does any power of, so I epect to dominate the function. Since as, I ll guess that the value for the it is. However, we must use L Hopital s Rule to be sure. 2

3 We re going to use a simple trick: rewrite ln ln as. Then the it results in the indeterminate form ln, we can now apply L Hopital s Rule: ln LR = = = 2 2 =. Indeterminate Differences If a it of quotients results in the form, adding any fractions appearing in the problem will often enable us to use L Hopital s Rule. Eample. In cot, rewriting cot as / tan allows us to combine the terms into a single fraction: Thus our it may be rewritten as cot = tan = tan tan. cot = tan tan ; since this second it has form, we are now free to use L Hopital s Rule: cot tan = tan 3 sec 2 tan sec 2.

4 Once again, we have form, so we try L Hopital s Rule again: sec 2 tan sec 2 2 sec 2 tan 2 sec 2 2 sec 2 tan = 2 =. Indeterminate Powers We cannot apply L Hopital s Rule directly to an indeterminate power such as, but we can use another trick to rewrite the epression so that L Hopital s Rule does apply. Eample. Find. Before working through the problem, let s again apply a bit of intuition. As, the base wants the function to go to, but the power wants the function to go to. My guess is that the power should dominate the base, i.e. the it will be. However, we must use L Hopital s Rule to check. Since evaluation gives us the form, we will rewrite using the eponential function and its inverse, the natural logarithm. Let s recall a few facts about these functions: first of all, e ln = whenever >. Second, ln a = a ln. Thus we rewrite Thus the original problem may be rewritten as ln = e = e ln. = e ln = e ln Thus the problem really amounts to calculating the inner it ln 4

5 This new it again gives us an an indeterminate form, this time the product ; so we rewrite ln = ln, and ln ln =. Since the it on the right now has form, L Hopital s Rule may be applied: ln 2 = =. We originally wanted to evaluate since ln =, we know that e ln ; = e ln = e ln = e =. So =. Forms that are not indeterminate The following forms are not indeterminate:,,,,,. All of the forms above have terms working towards a common goal; for instance, a it of the form is identically since the base is approaching, and raising very small numbers to increasingly large powers results in very small (close t 5