L'hopital's Rule - Suppose That f and g are Differentiable on an Open Interval I Containing

l’Hôpital’s Rule When we consider a limit which is a quotient of two functions, it is possible for us to encounter indeterminate forms, where the limit of both the numerator and denominator individually are either 0 or ∞. In such a situation we cannot apply the quotient rule for limits, but if we were to blindly apply the rule we would have something that looks like 0 0 ∞ , ∞ or which are both meaningless expressions. As we saw before, whether or not such a limit exists depends on the rate at which the functions in the numerator and denominator approach 0 or ∞. Using this notion we were able to characterize the order of magnitude of growth and decay of functions. At the time we did not have access to l’Hôpital’s rule, so we took such relationships for granted. However, with access to differentiation, we can actually evaluate limits of these types directly. Theorem 0.1.1 (l’Hôpital’s Rule). Suppose that f and g are differentiable on an open interval I containing x0 ∈ R∗ ≡ R ∪ {±∞}, and lim |f (x)| = lim |g(x)| = ∞ x→x0 x→x0 or lim f (x) = lim g(x) = 0. x→x0 x→x0 Further, suppose that g 0 (x) 6= 0 for all x 6= x0 in I. It follows that lim x→x0 f 0 (x) f (x) = lim 0 , g(x) x→x0 g (x) provided that the right limit exists, or is ±∞. When we say that I contains ∞ or −∞ we mean that I is either of the form (−∞, b) or (a, ∞)). This is useful if we have a limit of the form ex = lim x→∞ x x→∞ lim d x dx e d dx x ex = ∞, x→∞ 1 = lim where limx→∞ ex = limx→∞ x = ∞ allowed us to apply the theorem. This result tells us that ex grows faster than x as x → ∞. As before, this allows us to classify the rates at which functions grow/decay as as x → x0 , by seeing if the limit is 0, a nonzero constant, or ±∞. When the limit is 0, we say the denominator either grows on a higher order of magnitude, or decays on a lower order of magnitude than the numerator (depending on whether the functions are growing or decaying). If the limit is ±∞, then we say the denominator either grows on a lower order of magnitude, or decays on a higher order of magnitude. Finally, if the limit is a nonzero constant, we say the functions are growing or decaying on the same order of magnitude. This is a very useful and powerful tool when we want to think about the growth and decay of functions. We can even apply this theorem to functions where the derivative is not defined at the point of interest but the limit still makes sense, such as a function like ln(x). The natural logarithm and its derivative are not defined for x0 = 0, but we can still look at the limit as x → 0. For instance, ln(x) −x2 1/x = lim = lim = 0. x→0 1/x x→0 x x→0 −1/x2 lim 1 which tells us that as x → 0 we have ln(x) → −∞ on a lower order of magnitude than 1/x → ∞. Finally, we can use l’Hôpital’s rule multiple times successively. l’Hôpital’s rule tells us that if a limit is in an indeterminate form, we can evaluate the limit by looking at the derivatives of the functions involved in the limit, provided that the derivatives exist. If we are faced with another indeterminate form, we can once again apply l’Hôpital’s rule, provided that the second derivatives exist. In most cases, we should eventually find a limit that is not in an indeterminate form, which will tell us the limit of the original two functions. It should be emphasized that if a limit is not in an indeterminate form, then l’Hôpital’s rule cannot be applied. e3x − 1 . x→0 x Solution By inspection, we can see that both the numerator and denominator are differentiable, and the limit is of the indeterminate form 0/0 so it is okay to apply the rule. Thus, we find Example 1. Evaluate lim e3x − 1 3e3x = lim = 3. x→0 x→0 1 x lim e3x − 1 . Example 2. Evaluate lim x→∞ x Solution In this situation, we know that ex approaches ∞ faster than x as x → ∞. Thus, the limit should be ∞. Let us verify this using l’Hôpital’s rule. e3x − 1 3e3x = lim = ∞. x→∞ x→∞ 1 x lim e3x . x→0 x Solution In this case we do not have an indeterminate form, so we cannot apply l’Hôpital’s rule. As x → 0 the numerator approaches 1 while the denominator approaches 0. In this situation we can only conclude that e3x lim does not exist. x→0 x because looking at the one-sided limits we have Example 3. Evaluate lim e3x = −∞ and x%0 x lim e3x = ∞. x&0 x lim x2 − 1 . x→1 x − 1 Solution First notice that this limit is in the indeterminate form 00 , so we can apply l’Hôpital’s rule. x2 − 1 2x lim = lim =2 x→1 x − 1 x→1 1 Alternatively, we can write x2 − 1 = (x + 1)(x − 1), and cancel a factor of x − 1 from the numerator and denominator. Then, we find Example 4. Evaluate lim x2 − 1 = lim x + 1 = 2. x→1 x→1 x − 1 lim 3x − sin(x) . x→0 x Solution This limit is in the indeterminate form 00 , so we can apply l’Hôpital’s rule, finding Example 5. Evaluate lim lim x→0 3x − sin(x) 3 − cos(x) = lim = 2. x→0 x 1 2 √ 1 + x − 1 − x/2 . x2 Solution Once again, this limit is in the indeterminate form 00 , so we can apply l’Hôpital’s rule. Example 6. Evaluate lim x→0 √ lim x→0 (1/2)(1 + x)−1/2 − 1/2 −(1/4)(1 + x)−3/2 1 1 + x − 1 − x/2 = lim = lim =− . x→0 x→0 x2 2x 2 8 ln(x) √ . 2 x Solution Now we have an indeterminate form Example 7. Evaluate lim x→∞ ∞ ∞. We find ln(x) 1/x 1 √ = lim √ = lim √ = 0. x→∞ 2 x x→∞ 1/ x x→∞ x lim It is also possible for us to encounter indeterminate forms other than 0/0 and ∞/∞. For instance, we might be faced with forms such as 0 · ∞ or ∞ − ∞. In order to deal with such forms, we need to use some algebraic tricks to write them in the form of 0/0 or ∞/∞, and then we can apply l’Hôpital’s rule. Consider the following examples. 1 1  Example 8. Evaluate the limit lim . − x→0 x sin(x) Solution Here when we evaluate this limit we have two vanishing denominators, which leaves us with ∞ − ∞, a meaningless result. If we multiply these fractions to a common denominator, we find 1 x − sin(x) 1  = lim lim − x→0 x sin(x) x→0 x sin(x) The leaves us in the form of 0/0, so we can apply l’Hôpital’s rule. x − sin(x) 1 − cos(x) sin(x) 0 = lim = lim = = 0. x→0 x sin(x) x→0 sin(x) + x cos(x) x→0 2 cos(x) − x sin(x) 2 lim Example 9. Evaluate the limit lim tan(x) · ln(sin(x)). x→π/2 Solution Here we have a limit of the form ∞ · 0, but by simply writing tan(x) = 1/ cot(x) we have ln(sin(x)) lim tan(x) · ln(sin(x)) = lim , x→π/2 x→π/2 cot(x) which is of the form 0/0. Applying l’Hôpital’s rule lim x→π/2 ln(sin(x)) cos(x)/ sin(x) = lim = lim (− cos(x) sin(x)) = 0. cot(x) − csc2 (x) x→π/2 x→π/2 3