Lecture 28 Notes: Indeterminate Forms and L’Hospital’s Rule

18.01 Calculus Jason Starr Fall 2005 Lecture 28. December 1, 2005 Homework. Problem Set 8 Part I: (a) and (b). Practice Problems. Course Reader: 6A­1, 6A­2. 1. Indeterminate forms. Expressions of the form 0/0, ∞/∞, 0 × ∞, ∞ − ∞, 0∞ and ∞0 are called indeterminate forms. To be precise, none of these expressions is defined in mathematics. However, if a naive limit computation limx→a F (x) leads to an indeterminate form, it often happens that a more careful computation using calculus eliminates the indeterminate form. Example. Let b be any real number. Compute the limit as x approaches 0 of F (x) = (b+1/x)−1/x, x= � 0. If we evaluate this limit in a naive manner, we get, � � � � 1 1 1 1 − “=” lim b + − lim = ∞ − ∞. lim F (x) = lim b + x→0 x→0 x→0 x x x x→0 x This is an indeterminate form. In other words, the computation of the limit failed to give any useful information. The reason is that the general formula, lim [g (x) + h(x)] = lim g (x) − lim h(x), x→a x→a x→a only holds if all three limits are defined, which they are not in our case. Of course F (x) is simply the constant function with value b. Therefore, lim F (x) = lim b = b. x→0 x→0 Thus, a more careful computation proves the limit exists and gives its value. 2. The Mean Value Theorem revisited. Recall the Mean Value Theorem: If f (x) is continuous on [a, b] and differentiable on (a, b), then for some c strictly between a and b, f � (c) = f (b) − f (a) . b−a 18.01 Calculus Jason Starr Fall 2005 Thus, given two such functions f (x) and g(x) such that g(b) − g(a) is nonzero, there exist two values c1 and c2 strictly between a and b such that, f � (c1 ) (f (b) − f (a))/(b − a) f (b) − f (a) = = . g � (c2 ) (g(b) − g(a))/(b − a) g(b) − g(a) Is there a single value c = c1 = c2 where this equality holds? The answer is yes. Form the function F (x) = (f (b) − f (a))(g(x) − g(a)) − (g(b) − g(a))(f (x) − f (a)). Since f (x) and g(x) are continuous on [a, b], also F (x) is continuous on [a, b]. Since f (x) and g(x) are differentiable on (a, b), also F (x) is differentiable on (a, b). Moreover, F (a) = F (b) = 0. Thus, by the Mean Value Theorem, there exists a value c strictly between a and b such that F � (c) = 0. By a straightforward computation, F � (c) = (f (b) − f (a))g � (c) − (g(b) − g(a))f � (c). This proves the Generalized Mean Value Theorem. The main consequence of the Generalized Mean Value Theorem is the following result. Proposition. Let f (x) and g(x) be continuous functions on [a, b] that are differentiable on (a, b). If g � (x) is nonzero on (a, b), then g(x) − g(a) is nonzero for all a < x < b so that the expression, f (x) − f (a) g(x) − g(a) is defined. The right­handed limit, lim+ x→a f (x) − f (a) , g(x) − g(a) exists if and only if the right­handed limit, lim+ x→a f � (x) , g � (x) exists. If both limits exist, they are equal, lim+ x→a f � (x) f (x) − f (a) . = lim+ � x→a g (x) g(x) − g(a) A similar result holds for left­handed limits. The proof follows by applying the Generalized Mean Value Theorem to the interval [a, x] to replace (f (x) − f (a))/(g(x) − g(a)) by f � (c)/g � (c). Then x approaches a as c approaches a. 18.01 Calculus Jason Starr Fall 2005 3. L’Hospital’s rule. The most important case of the proposition is L’Hospital’s rule. This is exactly the case when f (a) = g(a) = 0. In this case, a naive computation would give, lim+ x→a f (a) 0 f (x) “=” = , g(x) g(a) 0 which is an indeterminate form. Again, the problem is that the general formula, lim+ x→a f (x) limx→a+ f (x) , = g(x) limx→a+ g(x) only holds if all three limits are defined, and the limit limx→a+ g(x) is nonzero. Since the limit is zero, the formula does not hold. However, if f � (x) and g � (x) exist, and if g � (x) is nonzero, then the proposition has the following consequence, known as L’Hospital’s rule, limx→a+ f (x)/g(x) = limx→a+ f � (x)/g � (x). Examples. sinh(x) cosh(x) 1 = lim = = 1. x→0 cos(x) x→0 sin(x) 1 lim 12x2 12 · 4 48 4x3 − 32 = lim = = = 16. 2 x→2 x − x − 2 x→2 2x − 1 2·2−1 3 1 − cos(x) lim = lim sin(x)2x = lim cos(x)2 = 1/2. x→0 x→0 x→0 x2 lim 4. L’Hospital’s rule for other indeterminate forms. L’Hospital’s rule can be used to compute limits that naively lead to indeterminate forms other than 0/0. For instance, if lim f (x) = lim+ g(x) = ∞, x→a+ x→a then the naive computation gives, lim+ x→a f (x) ∞ “=” . g(x) ∞ Now observe that, lim (1/f (x)) = lim+ (1/g(x)) = 0. x→a+ x→a � Therefore, if both g(x) and g (x) are nonzero on (a, b), then L’Hospital’s rule gives, lim+ x→a (1/f (x)) (1/f (x))� −f � (x)/f (x)2 = lim+ = lim . x→a+ −g � (x)/g(x)2 (1/g(x)) x→a (1/g(x))� 18.01 Calculus Jason Starr Fall 2005 Assuming that the limits, lim+ x→a f (x) f � (x) , and lim+ � x→a g (x) g(x) are defined and nonzero, the formula above can be re­written as, � �−1 � � � �−2 f � (x) f (x) f (x) · lim+ lim = lim+ � . x→a g (x) x→a g(x) x→a+ g(x) Solving gives, limx→a+ f (x)/g(x) = limx→a+ f � (x)/g � (x), if both limits are defined and nonzero. In fact, a better result is true (with a more subtle proof): if the second limit is defined, then the first limit is defined and the 2 are equal (whether or not they are zero). Example. lim + x→π/2 ln(x − π/2) 1/(x − π/2) = lim + = · · · = 0. x→π/2 sec(x) tan(x) sec(x) By similar arguments, other indeterminate forms can also be reduced to L’Hospital’s rule. Also, limits of the form, lim F (x) x→∞ giving indeterminate forms can often be reduced to L’Hospital’s rule. The moral is that the formula, f (x) f � (x) = lim � , x→a g(x) x→a g (x) lim is almost always true if f (a)/g(a) is an indeterminate form. But a certain amount of care should be used, since occasionally this fails.