l’Hôpital’s Rule In simple cases we are able to evaluate limits of the form lim x→x0 f (x) g(x) where limx→x0 f (x) = limx→x0 g(x) = ∞ or limx→x0 f (x) = limx→x0 g(x) = 0 using knowledge of the different rates at which functions approach ∞ or 0, respectively. However, if we are faced with a more complicated limit, such as e3x − 1 x→0 x lim it is not clear at what relative rates the numerator and denominator approach 0. When we are faced with a limit of 0 ∞ or 0 ∞ the limit is said to be in an indeterminate form. In general, to evalute limits of this type we will need to use l’Hôpital’s Rule. l’Hôpital’s Rule Suppose that f (x) and g(x) are differentiable on an open interval I containing x0 , and that lim f (x) = lim g(x) = ±∞ x→x0 x→x0 or lim f (x) = lim g(x) = 0 x→x0 x→x0 0 Further, suppose that g (x) 6= 0 for all x 6= x0 in I. It follows that lim x→x0 f (x) f 0 (x) = lim 0 g(x) x→x0 g (x) provided that the right limit exists, or is ±∞. Essentially, 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 indetermine form, then l’Hôpital’s rule cannot be applied. Example 1 Evalue the limit e3x − 1 x→0 x lim Solution As stated before, we will need to use l’Hôpital’s rule to evaluate this limit. By inspection, we can see that both the numerator and denominator are differentiable, so it is okay to apply the rule. Thus, we find e3x − 1 3e3x lim = lim =3 x→0 x→0 1 x Example 2 Evalue the limit e3x − 1 x→∞ x x Solution In this situation, we know that e approaches ∞ faster than x as x → ∞. Thus, the limit should be ∞. Let us verify this using l’Hôpital’s rule. lim e3x − 1 3e3x = lim =∞ x→∞ x→∞ 1 x lim Example 3 Evalue the limit e3x lim 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, so we find that e3x =∞ lim x→0 x Example 4 Evalue the limit x2 − 1 lim 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 x2 − 1 = lim x + 1 = 2 lim x→1 x→1 x − 1 Example 5 Evalue the limit 3x − sin(x) x→0 x Solution This limit is in the indeterminate form 00 , so we can apply l’Hôpital’s rule. lim 3x − sin(x) 3 − cos(x) = lim =2 x→0 x→0 x 1 lim Example 6 Evalue the limit √ lim x→0 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. √ 1 + x − 1 − x/2 (1/2)(1 + x)−1/2 − 1/2 −(1/4)(1 + x)−3/2 1 lim = lim = lim =− 2 x→0 x→0 x→0 x 2x 2 8 Example 7 Evalue the limit ln(x) √ x→∞ 2 x lim Solution Now we have an indeterminate form ∞ . ∞ We find ln(x) 1/x 1 √ = lim √ = lim √ = 0 x→∞ 2 x x→∞ 1/ x x→∞ x lim