3  L’Hôpital’s Rule

SageMath Spotlight

One of the most powerful ideas in calculus, which pervades all of mathematics, is the notion of linearization. Often, our problems are very hard—but lines are much easier!—and if we can find systematic ways to turn difficult questions into problems about lines, then life becomes much better. In Calculus I, we used limits to define derivatives. In this section, we use the power of derivatives—the power of linearization—to approach difficult questions about limits.

3.1 Indeterminate Forms

Notion of Indeterminate Forms (Stewart et al. 2020, 4.4, pp. 309)

A limit is considered an indeterminate form when it is built of continuous functions we understand, but they are in tension with one another in a way that makes deducing their limit non-obvious at first glance.

Example 3.1 The limit \[ \lim_{x \to 0} \frac{\sin x}{x} \] is an indeterminate form, because both functions (\(\sin x\) and \(x\)) tend to zero as \(x \to 0\). The shrinking quantity in the numerator is trying to make the limit very small, while the shrinking quantity in the denomoinator is trying to make the limit blow up to infinity.

For shorthand, this sort of limit is called a \(\frac{0}{0}\) indeterminate form, even though we never actually divide by \(0\) (remember, limits are about approaching a given value, not attaining it).

Seven Indeterminate Forms

In this course, we consider indeterminates of several related forms. We include each form here, together with two examples each which converge to different limits to help indicate why these forms are indeterminate:

Indeterminate Form Example 1 Example 2
\(\frac{0}{0}\) \(\displaystyle \lim_{x \to 0} \frac{x}{x} = 1\) \(\displaystyle \lim_{x \to 0} \frac{x^2}{x} = 0\)
\(\frac{\infty}{\infty}\) \(\displaystyle \lim_{x \to \infty} \frac{2x}{x+3} = 2\) \(\displaystyle \lim_{x \to \infty} \frac{e^x}{x} = \infty\)
\(0 \cdot \infty\) \(\displaystyle \lim_{x \to \infty} \frac{1}{x^2} \cdot e^x = \infty\) \(\displaystyle \lim_{x \to 0^+} x \ln(x) = 0\)
\(\infty - \infty\) \(\displaystyle \lim_{x \to \infty} (x+5) - x = 5\) \(\displaystyle \lim_{x \to \infty} x^3 - x = \infty\)
\(1^\infty\) \(\displaystyle \lim_{x \to \infty} 1^x = 1\) \(\displaystyle \lim_{x \to \infty} \left(1+\frac{1}{x}\right)^x = e\)
\(0^0\) \(\displaystyle \lim_{x \to 0^+} x^x = 1\) \(\displaystyle \lim_{x \to 0^+} x^{\frac{1}{\ln(x)}} = e\)
\(\infty^0\) \(\displaystyle \lim_{x \to \infty} x^{\frac{1}{x}} = 1\) \(\displaystyle \lim_{x \to \infty} (e^x)^{\frac{1}{\sqrt{x}}} = \infty\)

Limits of this forms are indeterminates because we cannot immediately deduce what their limit is (or even if such a limit exists)! In order to analyze them more carefully, we need a technique to scrutinize the tensions between the competing functions. Our main tool is:

L’Hôpital’s Rule (Stewart et al. 2020, 4.4, pp. 310)

Suppose \(f\) and \(g\) are differentiable and that \(g'(x) \not = 0\) on an interval containing \(a\) (except possibly at \(a\)). Suppose that \(\frac{f(x)}{g(x)}\) is of the indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), i.e., that \[ \lim_{x \to a} f(x) = 0 = \lim_{x \to a} g(x) \] or that \[ \lim_{x \to a} f(x) = \infty = \lim_{x \to a} g(x). \] Note that, when \(a = \pm \infty\), the limits are understood as limits at infinity. Then \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \] if the limit on the right hand side exists.

Example 3.2 Consider \({\displaystyle \lim_{x \to \infty}} \frac{\ln x}{\sqrt{x}}\). Both the numerator and denominator grow arbitrarily large as \(x \to \infty\), so we have an indeterminate form \(\frac{\infty}{\infty}\). We compute: \[ \begin{aligned} \lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} & = \lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{2\sqrt{x}}} & \qquad & \text{(L’Hôpital’s Rule)} \\ & = \lim_{x \to \infty} \frac{2\sqrt{x}}{x} & & \text{(simplify fractions)} \\ & = \lim_{x \to \infty} \frac{2}{\sqrt{x}} = 0. & & \text{(simplify/evaluate)} \end{aligned} \] We can verify this in SageMath (see Section 2.6.2; remember that ∞ is encoded as oo):

var('x')
limit(ln(x)/sqrt(x), x=oo)
0

3.2 Relating Indeterminate Forms

Notice that, at the moment, L’Hôpital’s Rule only allows us to address two of the seven indeterminate forms. We introduce several tricks for rendering some forms into others, so that we can apply L’Hôpital’s Rule to make sense of them. Our main trick involves properties of exponentials and logarithms, namely \[ e^{u+v} = e^u\ e^v \quad \text{and} \quad \ln(AB) = \ln A + \ln B. \]

Consider indeterminate forms, say with respect to some limit \(x \to a\).

  • If \(f(x) g(x)\) is an indeterminate form \(0 \cdot \infty\), then \(\frac{f(x)}{\frac{1}{g(x)}}\) and \(\frac{g(x)}{\frac{1}{f(x)}}\) will be indeterminate forms \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), and the limits of any one of these three forms (if they exist) are all equal.

  • If \(f(x) - g(x)\) is an indeterminate form \(\infty - \infty\), then \(\frac{e^{f(x)}}{e^{g(x)}}\) is an indeterminate form \(\frac{\infty}{\infty}\). Moreover, if the latter limit exists \[ \lim_{x \to a} \frac{e^{f(x)}}{e^{g(x)}} = L, \] and \(L > 0\), then \(\displaystyle \lim_{x \to a} f(x) - g(x) = \ln(L)\).

  • If \(f(x)^{g(x)}\) is an indeterminate form \(0^0\), \(1^\infty\) or \(\infty^0\), then \(g(x) \ln\left( f(x) \right)\) is an indeterminate form \(0 \cdot \infty\). Moreover, if the latter limit exists \[ \lim_{x \to a} g(x) \ln\left( f(x) \right) = L, \] then \(\displaystyle \lim_{x \to a} f(x)^{g(x)} = e^L\).