Calculus I Lesson 18: Limits at Infinity

  1. Warm Up
  2. Exam Questions
  3. Presentation 2 Reminder
  4. Limits at Infinity
  5. L’Hôpital’s Rule
    1. Other Indeterminate Forms
  6. Homework

Warm Up

A 100’ by 200’ yard is fenced off. The owner wishes to use part of the back of the property for a garden. She has 60 feet of fencing. Since the back is already fenced off, she only needs to fence off three sides. What is the area of the largest garden she can make?

Follow-up: what if she wishes to use a “corner”? In other words, if she needs to fence off two sides?

Exam Questions

Presentation 2 Reminder

Starting next week. Schedule?

Limits at Infinity

Previously, when we studied limits, we looked at what happens as a function approaches a particular point. It’s also important to understand the end-behavior of functions. This is also discussed using the language of limits:

\[\lim_{x \rightarrow \infty} f(x) = L\]

means that as $x$ gets larger, $f(x)$ gets closer and closer to $L$. For example, the function $f(x) = \dfrac{1}{x}$ approaches $0$ as $x \rightarrow \infty$:

Notice that on the right, the function gets closer and closer to the $y$-axis. If $\lim\limits_{x \rightarrow \infty} f(x) = L$, then the line $y = L$ is a horizontal asymptote of the function $f(x)$. Similarly, we can look at $\lim\limits_{x \rightarrow -\infty} f(x)$, and if this has a limit, then there is also a horizontal asymptote of $f(x)$.

Example: Find $\lim\limits_{x \rightarrow \infty} \dfrac{2x + 1}{x}$.

Notice that $\dfrac{2x + 1}{x} = \dfrac{2x}{x} + \dfrac{1}{x}$, or $2 + \dfrac{1}{x}$. Therefore, as $x \rightarrow \infty$, $\dfrac{2x+1}{x} \rightarrow 2 + 0$, which is $2$.

Example: Find $\lim\limits_{x \rightarrow \infty} \dfrac{2x - 1}{3x - 2}$.

Here there is a trick: divide the numerator and denominator by $x$:

\[\frac{2x - 1}{3x - 2} = \frac{ \frac{2x - 1}{x}}{\frac{3x - 2}{x}} = \frac{2 - \frac{1}{x}}{3-\frac{2}{x}}\]

So as $x \rightarrow \infty$, $\dfrac{2x - 1}{3x - 2} \rightarrow \frac{2 - 0}{3 - 0} = \frac{2}{3}$.

Exercises:

Find the following limits:

  1. $\lim\limits_{x \rightarrow \infty} \dfrac{x - 1}{x^2 - 1}$. (Hint: factor)
  2. $\lim\limits_{x \rightarrow \infty} \dfrac{x^2 + 1}{x^2 - 1}$. (Hint: divide the numerator and denominator by $x^2$.)
  3. $\lim\limits_{x \rightarrow \infty} \dfrac{\sin(x)}{x^2}$.
Check your answers
  1. $\dfrac{x-1}{(x+1)(x-1)}=\dfrac{1}{x+1}$. As $x \rightarrow \infty$, this approaches 0.
  2. $\dfrac{x^2+1}{x^2 - 1} = \dfrac{1 + \frac{1}{x^2}}{1 - \frac{1}{x^2}}$. As $x \rightarrow \infty$, this appraoches $1$.
  3. Notice that $\|\sin(x)\| \leq 1$. That means, for every $x$, $-\frac{1}{x^2} \leq \frac{\sin(x)}{x^2} \leq \frac{1}{x^2}$. As $x \rightarrow 0$, $\frac{1}{x^2}$ and $-\frac{1}{x^2}$ both approach 0, and so the Squeeze Theorem implies that this function approaches 0 as well.

Some other examples that should not be surprising:

Exercises:

Find the following limits:

  1. $\lim\limits_{x \rightarrow \infty} \cos(x^2)$.
  2. $\lim\limits_{x \rightarrow \infty} e^{-x^2}$.
Check your answers
  1. DNE. Since $\cos$ oscillates between $-1$ and $1$, even as $x \rightarrow \infty$, $\cos(x^2)$ does not settle down to one particular number.
  2. 0. Notice that as $x \rightarrow \infty$, $-x^2 \rightarrow -\infty$. So the exponent approaches $-\infty$. As $y \rightarrow -\infty$, $e^y \rightarrow 0$, so that means $e^{-x^2} \rightarrow 0$.

For more practice, please look at Exercises 261-270 from Section 4.6.

L’Hôpital’s Rule

Sometimes, when evaluating limits at infinity, we end up with an expression looking like $\frac{\infty}{\infty}$, which we cannot evaluate. Some expressions of that form may end up approaching a finite limit, like $\frac{2x+1}{x - 1} \rightarrow 2$. Some may end up approaching infinity, like $\frac{x^2+1}{x + 2}$. Some may end up not existing at all!

$\frac{\infty}{\infty}$ is called an indeterminate form for this reason: the form of the limit does not determine the value. But there is something we can do in this case!

Theorem (L’Hôpital’s Rule) Suppose $f$ and $g$ are differentiable functions, $\lim\limits_{x \rightarrow a} f(x) = \infty$, and $\lim\limits_{x \rightarrow a} g(x) = \infty$. Then

\[\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f^\prime(x)}{g^\prime(x)}\]

provided the right hand side exists or is $\pm \infty$. Note that this theorem applies in the case that $a = \pm \infty$ as well.

A note of caution: this is not the quotient rule! This just means to take the derivative of the numerator and the denominator, separately!

For example, as $x \rightarrow \infty$, both $\ln(x)$ and $x$ approach $\infty$. So,

\[\lim_{x \rightarrow \infty} \frac{\ln(x)}{x} = \lim_{x \rightarrow \infty} \frac{\frac{1}{x}}{1} = 0\]

Exercises:

Determine the following limits using l’Hôpital’s rule:

  1. $\lim\limits_{x \rightarrow \infty} \frac{x^2}{e^x}$
  2. $\lim\limits_{x \rightarrow \infty} \frac{e^x}{x^3}$
  3. $\lim\limits_{x \rightarrow \infty} \frac{e^x}{x^{100}}$
Check your answers
  1. Applying l'Hôpital's rule twice, we end up with $\frac{2}{e^x}$, which approaches $0$ as $x \rightarrow \infty$.
  2. Applying l'Hôpital's rule three times, we end up with $\frac{e^x}{6}$, which approaches $\infty$.
  3. Notice the pattern: applying l'Hôpital's rule over and over again, eventually the denominator will become a constant, and the numerator will still be $e^x$. $\frac{e^x}{k}$ approaches $\infty$ for any constant $k$.

Other Indeterminate Forms

There are several other indeterminate forms that l’Hôpital’s rule may apply to:

You can look at Section 4.8 in the textbook for more information on these other forms. But for now, I will only focus on the $\infty / \infty$, $0 / 0$ and $0 \cdot \infty$ forms. The $0 / 0$ case is very similar to $\infty / \infty$: take the derivative of the numerator and denominator, separately, and then check the limit again. So let’s look at an example of a limit of the form $0 \cdot \infty$:

\[\lim_{x \rightarrow 0^+} x \ln(x)\]

Here it is good to know that l’Hôpital’s rule can be used in the case of one-sided limits as well. But first, since we do not have a numerator and a denominator, we should instead turn this into a quotient: $x \ln(x) = \frac{\ln(x)}{1 / x}$, and now we can take the derivative of the numerator and denominator separately:

\[\begin{align} \lim_{x \rightarrow 0^+} x \ln(x) = &= \lim_{x \rightarrow 0^+} \frac{\ln(x)}{1 / x} \\ &= \lim_{x \rightarrow 0^+} \frac{1 / x}{-1 / x^2} \\ &= \lim_{x \rightarrow 0^+} \frac{-x^2}{x} \\ &= \lim_{x \rightarrow 0^+} -x = 0 \end{align}\]

Exercise: Find $\lim\limits_{x \rightarrow \infty} x^2 e^{-x}$.

Check your answer

First notice that as $x \rightarrow \infty$, $x^2 \rightarrow \infty$ and $e^{-x} \rightarrow 0$, so we can apply l'Hopital's rule.

Re-write $x^2 e^{-x}$ as $\frac{x^2}{e^x}$, and then apply l'Hopital's rule twice. We end up with $\frac{2}{e^x}$, which approaches 0 as $x \rightarrow \infty$.

Homework

  1. DeltaMath due Thursday.
  2. Written Homework due Monday, 11/17:
    • Section 4.6 #262, 264
    • Section 4.7 #320, 326, 332, 334
    • Section 4.8 #358, 374, 380, 388
  3. Presentation 2: starting next week.