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Calculus I: Activities

Mark A. Fitch

Section 2.7 Inderminate Forms in Limits

Standards
  • Evaluate a limit requiring L’Hôpital’s Rule
  • Evaluate a limit involving exponential indeterminate form
While we can prove limit properties for sums of functions both of which are infinite or products of functions both of which are infinite, we have avoided forms such as \(\infty-\infty\) and \(0/0\text{.}\) In this section we will illustrate the reason these are a problem and learn ways to handle these forms when they occur in limits.

Subsection 2.7.1 Variability of Indeterminate Forms

Use the illustrations in Figure 2.7.1 and Figure 2.7.2 to see why limits of the form \(0/0\) and \(1^\infty\) cannot be evaluated directly.

Instructions.

Change the parameter \(a\) in the function \(f(x)=\frac{e^{ax}-1}{x}\) to see the value \(\lim_{x \to 0} f(x)\text{.}\)
Figure 2.7.1. Illustration of Limit Form 0/0

Activity 2.7.1. Why 0/0 is not determined.

Using the interactive illustration above we will illustrate that the expression \(0/0\) in a limit can equal any real value. This will demonstrate that we cannot write the answer to a limit evaluation when we see \(0/0\text{.}\)
(a)
Evaluate \(\frac{e^{1 \cdot x}-1}{x}\) at \(x=0\text{.}\) Write down the expression you obtain.
(b)
Make sure the slider is at \(a=1\text{.}\) Based on the graph what does it appear the function value is at \(x=0\text{?}\) Note this would be the limit value.
(c)
Evaluate \(\frac{e^{2 \cdot x}-1}{x}\) at \(x=0\text{.}\) Write down the expression you obtain.
(d)
Move the slider to \(a=2\text{.}\) Based on the graph what does it appear the function value is at \(x=0\text{?}\) Note this would be the limit value.
(e)
Move the slider around. For \(a\) between 1 and 5 what values does the function appear to take on at \(x=0\text{?}\)
(f)
Based on this experiment why does the expression \(0/0\) not determine a specific answer?

Instructions.

Change the parameter \(a\) and \(b\) in the function \(f(x)=(1+ax)^{b/x}\) to see the value \(\lim_{x \to 0^+} f(x)\text{.}\)
Figure 2.7.2. Illustration of Limit Form \(1^\infty\)

Activity 2.7.2. Why \(1^\infty\) is not determined.

Using the interactive illustration above we will illustrate that the expression \(1^\infty\) in a limit can equal any real value. This will demonstrate that we cannot write the answer to a limit evaluation when we see \(1^\infty\text{.}\)
(a)
Evaluate \((1+1x)^{1/x}\) at \(x=0\text{.}\) Write down the expression you obtain.
(b)
Make sure the sliders are at \(a=1\) and \(b=1\text{.}\) Based on the graph what does it appear the function value is at \(x=0\text{?}\) Note this would be the limit value.
(c)
Evaluate \((1+2x)^{1/x}\) at \(x=0\text{.}\) Write down the expression you obtain.
(d)
Move the slider to \(a=2\text{.}\) Based on the graph what does it appear the function value is at \(x=0\text{?}\) Note this would be the limit value.
(e)
Move the slider around. For \(a\) between 1 and 3 what values does the function appear to take on at \(x=0\text{?}\)
(f)
Based on this experiment why does the expression \(1^\infty\) not determine a specific answer?

Subsection 2.7.2 Indeterminate Forms \(0/0\) and \(\infty/\infty\)

Table 2.7.3. Indeterminate Forms
\(\frac{0}{0}\)
\(\frac{\infty}{\infty}\)
\(\infty-\infty\)
\(0 \cdot \infty\)
\(0^0\)
\(1^\infty\)
\(\infty^0\)
All of the expressions above are indeterminate, that is, they do not determine a specific value like 5 or infinity. This section demonstrates techniques to evaluate limits that initially have these forms. The goal is to identify an appropriate method and execute it correctly.

Example 2.7.5. Example of Using L’Hôpital’s Rule.

Evaluate \(\lim_{x \to 0} \frac{e^x-1}{x}\text{.}\)
Solution.
\begin{align*} \lim_{x \to 0} \frac{e^{ax}-1}{x} = \amp \amp\amp \text{ recognize form } 0/0 \\ \lim_{x \to 0} \frac{ae^{ax}}{1} = \amp a \amp\amp \text{ used L'Hôpital's Rule} \end{align*}

Subsection 2.7.3 More Indeterminate Forms

Example 2.7.6. Example of \(0 \cdot \infty\).

Evaluate \(\lim_{x \to \pi/2^+} \left(x-\frac{\pi}{2}\right)\tan x\text{.}\)
Solution.
\begin{align*} \lim_{x \to \pi/2^+} \left(x-\frac{\pi}{2}\right)\tan x = \amp\amp\amp \text{ recognize form } 0 \cdot \infty\\ \lim_{x \to \pi/2^+} \frac{x-\frac{\pi}{2}}{\frac{1}{\tan x}} = \amp\amp\amp \text{ clever algebra}\\ \lim_{x \to \pi/2^+} \frac{x-\frac{\pi}{2}}{\cot x} = \amp\amp\amp \text{ trigonometric identity}\\ \lim_{x \to \pi/2^+} \frac{1}{-\csc^2 x} = \amp\amp\amp \text{ L'Hôpital's Rule} \\ \lim_{x \to \pi/2^+} -\sin^2 x = \amp -1. \end{align*}

Example 2.7.7. Example of \(\infty-\infty\).

Evaluate \(\lim_{\theta \to 0^-} \csc\theta - \cot\theta\text{.}\)
Solution.
\begin{align*} \lim_{\theta \to 0^-} \csc \theta - \cot \theta = \amp\amp\amp \text{ recognize form } \infty-\infty\\ \lim_{\theta \to 0^-} \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} = \amp\amp\amp \text{ trigonometric identity}\\ \lim_{\theta \to 0^-} \frac{1-\cos \theta}{\sin \theta} = \amp\amp\amp \text{ algebra}\\ \lim_{\theta \to 0^-} \frac{\sin \theta}{\cos \theta} = \amp 0.\amp\amp \text{ L'Hôpital's Rule} \end{align*}

Example 2.7.8. Example of an exponential form.

Evaluate \(\lim_{x \to \infty} \left( 1+\frac{1}{x} \right)^{x^2}\text{.}\)
Solution.
\begin{align*} \lim_{x \to \infty} \left( 1+\frac{1}{x} \right)^{x^2} = \amp\amp\amp \text{ recognize form } 1^\infty\\ \lim_{x \to \infty} e^{x^2\ln(1+1/x)} = \amp\amp\amp \text{ definition of } f(x)^{g(x)}\\ e^{\lim_{x \to \infty} x^2\ln(1+1/x)} \amp\amp\amp \text{ continuity allows substitution}\\ \lim_{x \to \infty} x^2\ln(1+1/x) = \amp \\ \lim_{x \to \infty} \frac{\ln(1+1/x)}{x^{-2}} = \amp\amp\amp \text{ clever algebra}\\ \lim_{x \to \infty} \frac{\frac{1}{1+1/x}(-x^{-2})}{-2x^{-3}} = \amp\amp\amp \text{L'Hôpital's Rule} \\ \lim_{x \to \infty} \frac{x}{2(1+1/x)} = \amp \infty \amp\amp \text{ algebra}. \\ \lim_{x \to \infty} x^2\ln(1+1/x) = \amp \infty. \\ \lim_{x \to \infty} \left( 1+\frac{1}{x} \right)^{x^2} = \amp \\ \lim_{x \to \infty} e^{x^2\ln(1+1/x)} = \amp \\ \lim_{u \to \infty} e^u = \amp \infty. \end{align*}
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