Simple "real world" l'Hôpital's rule problem?

Question

I am on a team which is writing a set of lecture notes for differential calculus.

I am using a format of "Break ground" which poses a problem, "Dig in" which develops the tools to solve the problem, and "Reinforce" which practices using the tools we just developed.

While using an abstract mathematical problem for the "Break ground" section is fine, some people on my team would rather that this problem be a "real application" whenever possible.

We have been trying to find such an example for l'Hôpital's rule for a while, but we have run into the following roadblocks:

1. Simple l'Hôpital's rule problems (which require only one differentiation) can seemingly all be solved by appealing to the definition of the derivative. So it is only when we apply l'Hôpital's rule twice that the method seems "necessary". However, such a problem seems too complicated for a "first brush" with l'Hôpital.

2. We have two "real world" examples of l'Hopital in action.

   a. One in deriving continuous interest as a limit of discretely compounded interest. This is, perhaps, better done using the definition of the exponential function.

   b. The other involves taking the limit of the velocity function of a falling object with air resistance, as air resistance goes to $0$. This is a great example, but there are at least two essential variables (time after the fall $t$, and the coefficient of air resistance $k$). The presence of more than one variable could be confusing (especially when you are thinking of velocity as a function of time, but then taking the limit with respect to $k$).

Does anyone have an example of l'Hopital's rule for which it is both necessary (or at least convenient), relatively straight forward , and is a "real world application"?

If no one can come up with something meeting all 3 requirements, I will probably settle for using the "abstract" problem

$$ \lim_{x \to 1} \frac{x^4-2x+1}{4x^4-15x+11} $$

and then motivate the idea of the rule by "zooming in" on the point $(1,0)$ (play with this graph), and welcoming the students to try to approximate the limit by the ratio of the tangent line approximations at $x=1$.

Answer 1

Here's a possible problem: The Rayleigh–Jeans Law for black body radiation at a wavelength $\lambda$ was given by $$B_{RJ}(\lambda) = \frac{K}{\lambda^4}$$ where $K$ is a constant (depending on temperature, speed of light, Boltzmann's constant, but that's not important). The problem is this clearly gives divergent energy output as wavelength approaches zero.

Plank's Law (in light of quantum mechanics) instead gives $$B_P(\lambda) = \frac{k_1}{\lambda^5}\frac{1}{e^{k_2/\lambda}-1}$$

for constants $k_1$, $k_2$. What happens to Plank's Law as wavelength approaches $0$?

Note: This is also one of those great problems you can come back to with Taylor Series and show that for large $\lambda$ (and more details about the constants above) the two agree to a first-order approximation.

Answer 2

In thermodynamics and information theory the function $-x\log x$ for $0 \le x \le 1$ occurs in the definition of entropy. Just Google "entropy" and see. But that formula is not really defined at $x = 0$. It is natural to declare its value at $x=0$ to be $\lim_{x\rightarrow 0^+} -x\log x$. A graph makes it very plausible that the limit is $0$. This can be justified with L'Hospital's rule once, and that one use of the rule does not look like the definition of the derivative.

If you let $y=1/x$ then that limit is the same as $\lim_{y\rightarrow \infty} \log(y)/y$, which is obviously $0$ without using L'Hospital's rule if you understand how slowly logarithms grow, although such understanding is normally gained by L'Hospital's rule in the first place, e.g., to show $\lim_{y\rightarrow \infty} \log(y)/y^c = 0$ (or equivalently, $\lim_{x \rightarrow 0^+} x^c\log x = 0$) for any $c>0$ by L'Hospital.

As a different application, it's important that exponential functions (with base greater than 1) ultimately always grow faster than power functions. That is, if $a > 1$ and $n$ is a positive integer then $\lim_{x \rightarrow \infty} a^x/x^n = \infty$ or equivalently $\lim_{x \rightarrow \infty} x^n/a^x = 0$. This can be shown by L'Hospital's rule $n$ times (or just one time with induction). The special case $a = e$ is probably safer to focus on than the general case, and shows up in Euler's beautiful integral formula $n! = \int_0^\infty x^ne^{-x}\,dx$; it's good to know that the integrand decays to $0$ for large $x$.

Answer 3

I personally don't see any reason to teach L'Hopital's rule at all until after understanding how to manipulate asymptotic expansions. Even Taylor series should come after asymptotic series as a special nice case. The reason is that asymptotic expansions can do everything that L'Hopital can do and more, including handling non-differentiable functions, and is more intuitive. $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} \def\wi{\subseteq} $

For example, $\lim_{ d \to 0 } \lfrac{f(x+d)-2f(x)+f(x-d)}{d^2} = f''(x)$ for any function $f$ and $x \in \operatorname{Dom}(f)$ such that $f$ is twice-differentiable at $x$. This (and its higher order generalizations) can be used to numerically estimate the derivatives of a function given its values. It may seem that this is perfect to illustrate the power of L'Hopital's rule, but in fact it illustrates how asymptotic expansions actually give more information.

First, let's see how L'Hopital's rule can be used:

  $\lim_{ d \to 0 } \lfrac{f(x+d)-2f(x)+f(x-d)}{d^2}$

  $= \lim_{ d \to 0 } \lfrac{f'(x+d)-f'(x-d)}{2d}$ [by L'Hopital because $f$ is differentiable in a neighbourhood of $x$]

  $= \frac{1}{2} \lim_{ d \to 0 } \lfrac{f'(x+d)-f'(x)}{d} + \frac{1}{2} \lim_{ d \to 0 } \lfrac{f'(x-d)-f'(x)}{-d}$ [because these two limits exist]

  $= \frac{1}{2} \lim_{ d \to 0 } \lfrac{f'(x+d)-f'(x)}{d} + \frac{1}{2} \lim_{ d \to 0 } \lfrac{f'(x+d)-f'(x)}{d}$

  $= f''(x)$ [by definition of derivative of $f'$ at $x$]

So we get the result, but actually much more can be said about the result if we look at the asymptotic expansion of $f$ at the given $x$ as follows:

  If $f(x+d) \in f(x) + a d + b \frac{d^2}{2} + o(d^2)$ as $d \to 0$ for some constants $a,b$:

    $\lfrac{f(x+d)-2f(x)+f(x-d)}{d^2} \in b + o(1)$ as $d \to 0$.

This is strictly stronger than the earlier result, because some $f$ are not even once differentiable in any neighbourhood around $x$ but still satisfy the asymptotic condition. For example if $f(t) = t+\lfrac{1}{\lfloor \frac1{t^2} \rfloor}$ for any real $t \ne 0$, and $f(0) = 0$, then as $t \to 0$ we can easily obtain $f(t) \in t+\lfrac{1}{\frac1{t^2}+O(1)}$ $\wi t+\lfrac{t^2}{1+O(t^2)}$ $\wi t+t^2(1+O(t^2))$ $\wi t+t^2+o(t^2)$ and so $\lfrac{f(t)+f(-t)}{t^2} \in 1+o(1)$.

Now even if we are not interested in "not nice" functions, we still can say even more than L'Hopital's rule, specifically for "even nicer" thrice differentiable functions. It can be proven (one form of Taylor's theorem) that:

  For any function $f$ and $x \in \operatorname{Dom}(f)$ such that $f$ is $n$ times differentiable at $x$:

    $f(x+d) \in \sum_{k=0}^n f^{(k)}(x) \frac{d^k}{k!} + o(d^n)$ as $d \to 0$.

So we immediately get:

  For any function $f$ and $x \in \operatorname{Dom}(f)$ such that $f$ is thrice differentiable at $x$:

    $\lfrac{f(x+d)-2f(x)+f(x-d)}{d^2} \in f''(x) + o(d)$ as $d \to 0$.

As well as:

  For any function $f$ and $x \in \operatorname{Dom}(f)$ such that $f$ is $4$ times differentiable at $x$:

    $\lfrac{f(x+d)-2f(x)+f(x-d)}{d^2} \in f''(x) + f^{(4)}(x) \frac{d^2}{12} + o(d^3)$ as $d \to 0$.

Notice further that the approximation is exact if $f$ is a cubic polynomial. All these precise results cannot be obtained by L'Hopital. Also don't forget that a direct use of L'Hopital will fail to simplify all but the simplest limits, and so typically some clever manipulation is necessary. The problem is how to derive those manipulations! One way is, unsurprisingly, by analyzing the asymptotic expansions...

Answer 4

This is a nice question, and I look forward to seeing people's answers. You suggest that if we want examples that can't be done by the definition of the derivative, we can use ones that require second derivatives. An additional possibility is a limit at infinity, although that might or might not work for your order of topics. Here's a simple real-world example.

You have some money, and two choices of what to invest it in. A share in Acme, Inc., costs \$7, and returns a dividend of \$1 per year. A share of Glutco costs \$30 and gives a dividend of \$2 per year. If we want to compare the long-term value of the two investments, a natural way to do it is with the limit

$$ \lim_{t\rightarrow\infty}\frac{-7+t}{-30+2t} \quad .$$

The top represents the net return on Acme, the bottom Glutco. If this limit is greater than 1, then Acme is the better long-term investment. The constant terms are irrelevant, and the competition between the top and bottom is determined by which one grows faster.

Answer 5

This suggestion uses integrals and the fundamental theorem of calculus. If that is not yet known to the students, my example will not fit so well.

Background: The story says that Dirichlet (in a note written on a paper) around 1838 suggested to Gauss that the number of primes less than or equal to $x$, usually denoted $\pi(x)$, is for large $x$ approximately given by the integral $$ \int_2^x\frac{1}{\ln t}\,dt. $$ On the other hand, the prime number theorem states that $$ \lim_{x\to+\infty}\frac{\pi(x)}{x/\ln x}=1. $$

Exercise: Show that the integral suggested by Dirichlet is indeed a good approximation, in the sense that $$ \lim_{x\to+\infty}\frac{\ln x}{x}\int_2^x\frac{1}{\ln t}\,dt=1. $$

Comment: This limit appeared on math.SE, and it is indeed also solved easily using integration by parts. Nevertheless, I think it is a nice application of the l'Hospital rule.

Answer 6

I have a colleague whose hobby is to always figure out how to a evaluate any given limit without L'Hospital's rule. Every week in the lunch room, there is yet another example.

I think a very good application of the rule is to $\lim_{x\to 0}\frac{\sin(x)}{x}$.

I frequently teach the calculus sequence at the university level. And I am fortunate to have a daughter in high school who took AP Calculus AB as a junior and is now taking BC as a senior. On occasion I have mentioned L'Hospital to her. It has been interesting to gauge her reaction to the rule over time. For awhile a fake quotient rule took hold: $$\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)}{g'(x)}. $$ And she would resist paying attention to the conceptual derivation that involves linearization of the numerator and denominator. But as she becomes more comfortable with all the concepts, she appreciates this viewpoint and is even becoming a clone of my senior colleague. She often has the patience to understand how to evaluate limits without the rule, but at the same time feels confident using it when appropriate. And now she even appreciates the Taylor polynomial perspective.

What I think I have learned by observation is that the L'Hospital rule should be viewed as a "developmental" concept that threads through Calculus I and II, and the instructor (or parent!) needs to be patient.

Answer 7

In statistics we use the Box-Cox transform $$ y^{(\lambda)} = \begin{cases} \frac{y^\lambda -1}{\lambda}, & \text{if $\lambda \not= 0$} \\ \ln y, & \text{if $\lambda=0$} \end{cases} $$ as a continuous extension of the power transforms that include the natural logarithms as a limiting case when $\lambda=0$. To show the limit when $\lambda \rightarrow 0$ use L'hopitals rule.

Another example from statistics is the following. The $\alpha$-divergence between two probability densities defined on the same space is $$ d_\alpha (f || g) = \frac1{\alpha-1} \ln \int f(x)^\alpha g(x)^{1-\alpha}\; dx $$ The limit when $\alpha \rightarrow 1$ is the Kullback-Leibler divergence: $$ D_{KL}(f || g) = \int f(x) \ln\frac{f(x)}{g(x)} \; dx $$ which can be calculated by L'hopital.

Answer 8

This one is somewhat "real-world" in that it involves a question that can be posed without knowing anything about L'Hopital's rule. But I must warn you that it takes three iterations.

From the point $(\cos\theta,\sin\theta)$ on the unit circle, drop a perpendicular to the $x$-axis, and consider the region bounded by that perpendicular, the $x$-axis, and the arc from $(1,0)$ back to $(\cos\theta,\sin\theta)$. Ask what fraction of the area of that region is between the chord from $(1,0)$ to $(\cos\theta,\sin\theta)$ and the arc from $(1,0)$ to $(\cos\theta,\sin\theta)$, and then take the limit of that fraction as $\theta\downarrow0.$

Answer 9

Suppose a company has two new types of sweet to test out. The company gets two jars of the sweets and a lot of children. They allow the children to eat the sweets and frequently weigh the jars to see which sweet is most popular. The company will sell packets of sweets according to the ratio the children eat the sweets in. As time goes on the ratio of the weights will go towards $\frac{0}{0}$ and can not be computed. A solution is to compare the rate of change of each weight between measurements.

Answer 10

At the beginning of a race, the race cars have zero position and velocity, but they will have some non-zero derivative. So have a red car with greater acceleration and lower top speed, and a blue car with lower acceleration and greater top speed, and graph the ratio of distance travelled as a function of time. See if the ratio at early times depends on top speed or what. The position vs time functions can be chosen for convenience (your polynomials are probably good enough if you make a few simple modifications to give zero initial speed and velocity and nonzero acceleration.)

I think the benefit of this is that they are working with concepts they are familiar with (beginning a race instead of $x=0$, starting at the beginning from rest instead of $f(0)=0,\ f'(0)=0$, acceleration instead of $f''(x)$, etc).