Differential Calculus: From Practice to Theory

Based on our experience so far it is tempting to conclude that we can locate vertical asymptotes by simply setting the denominator equal to zero and solving. But this is a naive conclusion as we will see next.

We were careful to use one-sided limits to find the vertical asymptote at \(x= -2\) because this is what Definition 12.3.1 calls for. Strictly speaking we only needed one of them, but since these limits were not equal we evaluated them both.

Observe that this limit has the form \(\limit{x}{2}{\frac{x-2}{x^2-4}}=\frac{\approach{0}}{\approach{0}}\text{.}\) A limit of this form is another indeterminate Form like \(\approach{\infty -\infty }\text{.}\)

Example 12.4.1 demonstrates that a function which does not exist at a point is a more general concept than a function with a vertical asymptote at a point.

Do you see the difference? The value of \(g(2)\) is not defined because \(g(2)=\frac{2-2}{4-4}\) and we get a zero denominator. But no such difficulty occurs with \(f\) because \(f(2) = \frac{1}{2+2}=\frac14\text{.}\)

So, when \(x\neq2\text{,}\) \(g(x)\) and \(f(x)\) are the exactly the same. But they disagree when \(x=2\text{.}\) More precisely \(x=2\) is not in the domain of \(g(x)\) but it is in the domain of \(f(x)\text{.}\) Since the domains differ they must be different functions and it is incorrect to write \(\frac{x-2}{x^2-4}=\frac{1}{x+2}\) even though it is true for all but one value of \(x\text{.}\)

We found earlier that \(\tlimit{x}{\infty}{\frac{\sin(x)}{x}}=0\text{,}\) so the function \(y=\frac{\sin(x)}{x}\) must have a horizontal asymptote at \(y=0\text{.}\) Clearly \(y=\frac{\sin(x)}{x}\) is undefined at \(x=0\text{.}\) Does its graph have a vertical asymptote at \(x=0\text{?}\) Take your best guess.

To answer this question we need to evaluate the limit \(\tlimit{x}{0}{\frac{\sin(x)}{x}}\text{.}\) But how can we do that? If we try the obvious ploy of just “plugging in” zero for \(x\) we get the indeterminate form \(\frac{(\rightarrow0)}{(\rightarrow0)}\text{,}\) and we know we need to proceed cautiously with an indeterminate form.

From the graph of \(\frac{\sin(x) }{x}\) (in the sketch below) it would appear that \(\frac{\sin(x)}{x}\rightarrow1\) as \(x\rightarrow0\text{.}\) This is correct, but we’ve learned to be wary of relying on graphs. It would be best to evaluate the limit by another technique if possible.

Can we apply some algebraic trick like we did in Example 12.4.1?

It is hard to see what it would be. There isn’t much algebra we can do here. If there were other trigonometric functions involved we might be able to employ some trigonometric identity in the numerator, but there just isn’t much we can do with \(\sin(x)\text{.}\) This problem poses a bit more of a challenge than the indeterminate forms we have dealt with so far. We need a new idea.

Since the \(\sin(x)\) isn’t helping maybe we can replace it with something simpler. Seriously. Give some thought to what might be a valid replacement before reading on. Remember that we’re really only interested in values of \(x\) that are very close (infinitely close?) to zero.

The Principle of Local Linearity says that the graph of a function and its tangent line are locally indistinguishable. Since we’re only interested in values of \(x\) near zero — locally around zero — let’s try replacing \(\sin(x)\) with its tangent line.

Let’s try this again with \(\tlimit{x}{2}{\frac{x^3-8}{x^2-4}}\) which also has the indeterminate form \(\frac{\approach{0}}{\approach{0}}\text{.}\)

In all of our examples where the indeterminate form was a fraction, replacing the numerator and denominator with their respective tangent lines seems to have worked. The idea seems promising so let’s look at it in a more general context. If \(f(x)\) and \(g(x)\) are differentiable functions such that \(f(a)=g(a)=0\) then \(\limit{x}{a}{\frac{f(x)}{g(x)}}\) is an indeterminate form.

This idea is due to the Swiss mathematician Johann Bernoulli

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https://mathshistory.st-andrews.ac.uk/Biographies/Bernoulli_Johann/

(1667–1748). Johann and his brother Jacob Bernoulli

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https://mathshistory.st-andrews.ac.uk/Biographies/Bernoulli_Jacob/

(1665–1795) were taught Calculus by Leibniz himself, but soon showed themselves to be his equal, or nearly so.

The Marquis de L’Hôpital, Guillaume François Antoine de L’Hôpital’

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https://mathshistory.st-andrews.ac.uk/Biographies/De_LHopital/

was keenly interested in the new Calculus so he struck a deal with Johann. The Marquis agreed to pay Bernoulli to tutor him and to keep L’Hôpital abreast of new mathematical developments. This was done with the understanding that L’Hôpital would include some of Bernoulli’s research results in the Calculus textbook L’Hôpital was writing at the time. This turned out to be the very first Calculus textbook: Analyse des Infiniment Petits pour l’Intelligence des Lignes Courbes (Analysis of the Infinitely Small for the Understanding of Curved Lines)

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https://old.maa.org/press/periodicals/convergence/mathematical-treasure-l-hospital-s-analyse-des-infiniment-petits

. It was published in 1696.

Even though L’Hôpital acknowledged his indebtedness to Johann Bernoulli for this procedure it has always been known as L’Hôpital’s Rule.

As a mathematician L’Hôpital was not in the same league as Leibniz, Newton, or the Bernoullis — few are — but he was a very competent mathematician and his work helped to disseminate the ideas of Calculus just as Maria Gaëtana Agnesi’s text would do in the following century.

Now look back at the examples in this section and notice the following fact: In each case when the limit was finally evaluated it turned out to be the slope of the line tangent to the graph of the numerator divided by the slope of the line tangent to the graph of the denominator. But the slope of the line tangent to \(f(x)\) at \(x=a\) is simply \(\eval{\dfdx{y}{x} }{x}{a} = f^\prime(a) \text{.}\)

In light of equation (12.5) it appears that we don’t need to go to the trouble of finding the tangent lines. We can evaluate these indeterminate forms as follows:

Because L’Hôpital’s Rule applies directly to the indeterminate form \(\frac{(\rightarrow0)}{(\rightarrow0)}\) we will refer to this as a L’Hôpital Indeterminate. Not all indeterminate forms are L’Hôpital Indeterminates.

We want to emphasize very strongly that L’Hôpital’s Rule requires a L’Hôpital indeterminate. If we don’t have a L’Hôpital indeterminate we can not use L’Hôpital’s Rule. If you try to use L’Hôpital’s Rule when it does not apply, you will probably not get the correct answer. The difficulty here is that nothing in the computations will alert you that anything has gone wrong. You have to be watchful for this potential pitfall.

This limit is indeterminate of the form \(\frac{(\rightarrow0)}{(\rightarrow0)}\text{,}\) so it seems like L’Hôpital’s Rule would apply. But both versions of L’Hôpital’s Rule (so far) require that \(x\rightarrow a,\) where \(a\) is a real number. And of course, infinity still insists on not being a number, real or otherwise. If we want to use L’Hôpital’s Rule we’ll have to find a way to modify this limit so that L’Hôpital’s Rule does apply.

If we make the substitution \(z=\frac{1}{x}\) then as \(x\rightarrow\infty\text{,}\) \(\underset{z>0}{z\rightarrow0}\text{.}\)

Far more important than the answer in this example is the realization that we can use the same trick to extend L’Hôpital’s Rule to the case where \(x\) approaches both \(\infty,\) and \(-\infty.\)

Although we have not discussed it here, L’Hôpital’s Rule also applies directly to the indeterminate form \(\frac{\approach{\pm\infty}}{\approach{\pm\infty}}\text{,}\) as well as \(\frac{\approach{0}}{\approach{0}}\text{.}\) So we will also call these L’Hôpital Indeterminates.

Putting this all together gives us the general statement of L’Hôpital’s Rule:

Colloquially, the phrase “exponential growth” just means that something is growing very fast, but it has a more precise meaning in mathematics. It means the growth rate is described by an exponential function. For example neither \(f(x)=x^2\) nor \(f(x)=x^{200}\) grows exponentially. They are both said to grow “polynomially”. But \(f(x) = e^x\text{,}\) \(f(x)=2^x\text{,}\) and even \(f(x)=1.0001^x\) all do grow exponentially The next problem explores how “exponential growth” compares “polynomial growth.”

Did you guess correctly?

Notice that the trick we used was a simple substitution much like the ones we have used in the past to make things “easier on the eyes.” The difference here is that we’re not making anything “easier on the eyes.” If anything we’re making things harder to look at since re–writing \(x\) as \(\frac{1}{\frac1x}\) certainly doesn’t seem like it’s going to help much. Until we try it. Always try your ideas out no matter how crazy they seem to be. Only after you try an idea can you tell if it will work or not.

Recall that in Section 8.2 where we defined the effective yield of an investment which is compounded continuously. We were led to examine the expression \(\left(1+\frac{1}{m}\right)^m\) for very large values of \(m\text{.}\)

As a result of that investigation we accumulated considerable numerical evidence that in one year an investment of \(1\) growing at a nominal rate of \(5\%\text{,}\) compounded continuously would grow to \(e^{0.05}\approx 1.0513\text{,}\) but we were unable to do more than gather evidence. It was pretty convincing evidence but it was not proof since we didn’t yet have the mathematical technique necessary to evaluate the expression \(\left(1+\frac{1}{m}\right)^m\) as \(m\rightarrow\infty\text{.}\) Now we do.

Of these, only the last two are L’Hôpital Indeterminate forms. The others must be evaluated by some method other than directly applying L’Hôpital’s Rule. Often this simply means algebraically rearranging them into a L’Hôpital Indeterminate.