l’Hˆopital’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 0 ∞ Consider the limit sin(x) lim x→0 x If we look at the individual limits of the numerator and denominator they are both 0, yet we know this limit is 1. Previously we used numerical arguments to establish this limit, but it would be preferable to have a more general way of calculating such limits. We can develop a more general strategy for doing so, if we think about the rates at which the functions are approaching 0. If the numerator approaches 0 much faster than the denominator, then it stands to reason the overall limit should be zero. If the denominator approaches more quickly, then it seems as though the limit should be unbounded, ±∞.

Let’s think about two continuous functions, f(x) and g(x), where the limit

f(x) lim x→x0 g(x)

0 is of the form 0 (note that we do not use an equal sign, because we cannot evaluate the limit of a quotient as a quotient of the limits unless the limit in the denominator is nonzero). Now 0 since we want to consider the rate of change of these functions, also let the derivatives f (x0) 0 and g (x0) exist. Following the observation about the derivatives, let’s try and simplify a statement about the ratio of the derivatives, to deduce something about the ratio of the functions themselves.

0 f(x)−f(x0) f(x)−f(x0) f (x ) limx→x0 f(x) − f(x ) f(x) 0 = x−x0 = lim x−x0 = lim 0 = lim 0 g(x)−g(x0) x→x g(x)−g(x0) x→x x→x g (x0) limx→x 0 0 g(x) − g(x0) 0 g(x) 0 x−x0 x−x0 After the above analysis, it turns out that were able to show that knowledge about the derivatives of the functions in question allows us to evaluate the limit that is an indeterminate form. If we apply this to the function sin(x)/x we find

sin(x) cos(0) lim = = 1 x→0 x 1 which was much simpler than trying to evaluate the limit without the use of the derivative. This ‘rule’ that we have found for evaluating these limits is called l’Hˆopital’s rule. However, the full form of l’Hˆopital’s rule is actually much more powerful, and this is really a special case of it. We can call this the weak form of l’Hˆopital’s rule. The full-powered form follows. l’Hˆopital’s Rule ∗ Let x0 ∈ R ≡ R ∪ {±∞}. 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 f(x) f 0(x) lim = lim 0 x→x0 g(x) x→x0 g (x)

provided that the right limit exists, or is ±∞.

There are a handful of reasons why this theorem is much more powerful than the one we proved. First, we allow x0 = ±∞ (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 d ex ex lim = lim dx = lim = ∞ x→∞ x x→∞ d x→∞ 1 dx x

x where limx→∞ e = limx→∞ x = ∞ allowed us to apply the theorem. This result tells us that ex grows faster than x as x → ∞. In this way we can compare the rates at which functions grow 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 more quickly, or decreases more slowly than the numerator (depending on whether the functions are growing or decaying). If the limit is ±∞, then we say the denominator either grows more slowly, or decays more quickly. Finally, if the limit is a nonzero constant, we say the functions are growing or decaying at the same rate. This is a very useful and powerful tool when we want to think about the growth and decay of functions.

Second, we can 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) 1/x −x2 lim = lim = lim = 0 x→0 1/x x→0 −1/x2 x→0 x which tells us that as x → 0 we have ln(x) → −∞ slower than 1/x → ∞. Finally, we can use this form of l’Hˆopital’s rule multiple times successively. l’Hˆopital’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’Hˆopital’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 em- phasized that if a limit is not in an indetermine form, then l’Hˆopital’s rule cannot be applied.

Example 1 Evalue the limit e3x − 1 lim x→0 x Solution By inspection, we can see that both the numerator and denominator are differen- tiable, and the limit is of the indeterminate form 0/0 so it is okay to apply the rule. Thus, we find e3x − 1 3e3x lim = lim = 3 x→0 x x→0 1 Example 2 Evalue the limit e3x − 1 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’Hˆopital’s rule.

e3x − 1 3e3x lim = lim = ∞ x→∞ x x→∞ 1 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’Hˆopital’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 0 Solution First notice that this limit is in the indeterminate form 0 , so we can apply l’Hˆopital’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 = lim x + 1 = 2 x→1 x − 1 x→1 Example 5 Evalue the limit 3x − sin(x) lim x→0 x 0 Solution This limit is in the indeterminate form 0 , so we can apply l’Hˆopital’s rule. 3x − sin(x) 3 − cos(x) lim = lim = 2 x→0 x x→0 1 Example 6 Evalue the limit √ 1 + x − 1 − x/2 lim x→0 x2 0 Solution Once again, this limit is in the indeterminate form 0 , so we can apply l’Hˆopital’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 = − x→0 x2 x→0 2x x→0 2 8 Example 7 Evalue the limit ln(x) lim √ x→∞ 2 x ∞ Solution Now we have an indeterminate form ∞ . We find ln(x) 1/x 1 lim √ = lim √ = lim √ = 0 x→∞ 2 x x→∞ 1/ x x→∞ x 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 algebratic tricks to write them in the form of 0/0 or ∞/∞, and then we can apply l’Hˆopital’s rule. Consider the following examples.

Example 8 Evaluate the limit  1 1  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 1  x − sin(x) 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’Hˆopital’s rule. x − sin(x) 1 − cos(x) sin(x) 0 lim = lim = lim = = 0 x→0 x sin(x) x→0 sin(x) + x cos(x) x→0 2 cos(x) − x sin(x) 2 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’Hˆopital’s rule

ln(sin(x)) cos(x)/ sin(x) lim = lim = lim (− cos(x) sin(x)) = 0 x→π/2 cot(x) x→π/2 − csc2(x) x→π/2