# Calculus and Abbott and Costello

Although $n + 30 > n$ for all $n$, it’s also true that

$\displaystyle \lim_{n \to \infty} \frac{n+30}{n} = 1$.

That’s the subtle mathematical premise behind this classic comedy routine from Abbott and Costello. (This routine was the basis of a recent article in The College Mathematics Journal.)

# Different Ways of Computing a Limit: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on different ways of computing the limit

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}$

Part 1: Algebra

Part 2: L’Hopital’s Rule

Part 3: Trigonometric substitution

Part 4: Geometry

Part 5: Geometry again

# Different ways of computing a limit (Part 5)

One of my colleagues placed the following problem on an exam for his Calculus II course…

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}$

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #5. Another geometric approach. The numbers $x$ and $\sqrt{x^2+1}$ can be viewed as two sides of a right triangle with legs $1$ and $x$ and hypotenuse $\sqrt{x^2+1}$. Therefore, the length of the hypotenuse must be larger than the length of one leg but less than the sum of the lengths of the two legs. In other words,

$x < \sqrt{x^2+1} < x+1$,

or

$1 < \displaystyle \frac{\sqrt{x^2+1}}{x} < \displaystyle 1+\frac{1}{x}$.

Clearly $\displaystyle \lim_{x \to \infty} 1 = 1$ and $\displaystyle \lim_{x \to \infty} \left( 1 + \frac{1}{x} \right) = 1$. Therefore, by the Sandwich Theorem, we can conclude that $\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = 1$.

# Different ways of computing a limit (Part 4)

One of my colleagues placed the following problem on an exam for his Calculus II course…

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}$

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #4. The geometric approach. The numbers $x$ and $\sqrt{x^2+1}$ can be viewed as two sides of a right triangle with legs $1$ and $x$ and hypotenuse $\sqrt{x^2+1}$. So as $x$ gets larger and larger, the longer leg $x$ will get closer and closer in length to the length of the hypotenuse. Therefore, the ratio of the length of the hypotenuse to the length of the longer leg must be 1.

# Different ways of computing a limit (Part 3)

One of my colleagues placed the following problem on an exam for his Calculus II course…

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}$

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #3. A trigonometric identity. When we see $\sqrt{x^2+1}$ inside of an integral, one kneejerk reaction is to try the trigonometric substitution $x = \tan \theta$. So let’s use this here. Also, since $x \to \infty$, we can change the limit to be $\theta \to \pi/2$:

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = \displaystyle \lim_{\theta \to \pi/2} \frac{\sqrt{\tan^2 \theta+1}}{\tan \theta}$

$= \displaystyle \lim_{\theta \to \pi/2} \frac{\sqrt{\sec^2 \theta}}{\tan \theta}$

$= \displaystyle \lim_{\theta \to \pi/2} \frac{ \sec \theta}{\tan \theta}$

$= \displaystyle \lim_{\theta \to \pi/2} \frac{ ~~\displaystyle \frac{1}{\cos \theta} ~~}{ ~~ \displaystyle \frac{\sin \theta}{\cos \theta} ~~ }$

$= \displaystyle \lim_{\theta \to \pi/2} \frac{ 1}{\sin \theta}$

$= 1$.

# Different ways of computing a limit (Part 2)

One of my colleagues placed the following problem on an exam for his Calculus II course…

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}$

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #2. Using L’Hopital’s Rule. The limit has the indeterminant form $\infty/\infty$, and so I can differentiate the top and the bottom with respect to $x$:

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = \displaystyle \lim_{x \to \infty} \frac{ \displaystyle \frac{d}{dx} \left( \sqrt{x^2+1} \right) }{\displaystyle \frac{d}{dx} \left( x \right)}$

$= \displaystyle \lim_{x \to \infty} \frac{ \displaystyle \frac{1}{2} \left( x^2+1 \right)^{-1/2} \cdot 2x }{1}$

$= \displaystyle \lim_{x \to \infty} \frac{x}{\sqrt{x^2+1}}$.

Oops… it looks like I just got the reciprocal of the original limit! Indeed, if I use L’Hopital’s Rule again, I’ll just return back to the original limit.

So that doesn’t look very helpful… except it is. If I define the value of this limit to be equal to $L$, then I’ve just shown that $L = 1/L$ (assuming that the limit exists in the first place, of course). That means that $L = 1$ or $L = -1$. Well, clearly the limit of this nonnegative function can’t be negative, and so we conclude that the limit is equal to $1$.

# Different ways of computing a limit (Part 1)

One of my colleagues placed the following problem on an exam for his Calculus II course…

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}$

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #1. The straightforward approach, using only algebra:

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = \displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{\sqrt{x^2}}$

$= \displaystyle \lim_{x \to \infty} \sqrt{1 + \frac{1}{x^2}}$

$= \sqrt{1 + 0}$

$= 1$.

# Different definitions of e (Part 12): Numerical computation

In this series of posts, we have seen that the number $e$ can be thought about in three different ways.

1. $e$ defines a region of area 1 under the hyperbola $y = 1/x$.2. We have the limits

$e = \displaystyle \lim_{h \to 0} (1+h)^{1/h} = \displaystyle \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n$.

These limits form the logical basis for the continuous compound interest formula.

3. We have also shown that $\frac{d}{dx} \left(e^x \right) = e^x$. From this derivative, the Taylor series expansion for $e^x$ about $x = 0$ can be computed:

$e^x = \displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$

Therefore, we can let $x = 1$ to find $e$:

$e = \displaystyle \sum_{n=0}^\infty \frac{1}{n!} = \displaystyle 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \dots$

In yesterday’s post, I showed that using the original definition (in terms of an area under a hyperbola) does not lend itself well to numerically approximating $e$. Let’s now look at the other two methods.

2. The limit $e = \displaystyle \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n$ gives a somewhat more tractable way of approximating $e$, at least with a modern calculator. However, you can probably imagine the fun of trying to use this formula without a calculator.

3. The best way to compute $e$ (or, in general, $e^x$) is with Taylor series. The fractions $\frac{1}{n!}$ get very small very quickly, leading to rapid convergence. Indeed, with only terms up to $1/6!$, this approximation beats the above approximation with $n = 1000$. Adding just two extra terms comes close to matching the accuracy of the above limit when $n = 1,000,000$.

More about approximating $e^x$ via Taylor series can be found in my previous post.

# Different definitions of e (Part 10): Connecting the two definitions

In this series of posts, I consider how two different definitions of the number $e$ are related to each other. The number $e$ is usually introduced at two different places in the mathematics curriculum:

1. Algebra II/Precalculus: If $P$ dollars are invested at interest rate $r$ for $t$ years with continuous compound interest, then the amount of money after $t$ years is $A = Pe^{rt}$.
2. Calculus: The number $e$ is defined to be the number so that the area under the curve $y = 1/x$ from $x = 1$ to $x = e$ is equal to $1$, so that

$\displaystyle \int_1^e \frac{dx}{x} = 1$.

These two definitions appear to be very, very different. One deals with making money. The other deals with the area under a hyperbola. Amazingly, these two definitions are related to each other. In this series of posts, I’ll discuss the connection between the two.

In yesterday’s post, I proved the following theorem, thus completing a long train of argument that began with the second definition of $e$ and ending with a critical step in the derivation of the continuous compound interest formula. Today, I present an alternate proof of the theorem using L’Hopital’s rule.

Theorem. $\displaystyle \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt} = P e^{rt}$.

Proof #2. Let’s write the left-hand side as

$L = \displaystyle \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt}$.

Let’s take the natural logarithm of both sides:

$\ln L = \displaystyle \ln \left[ \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt} \right]$

Since $g(x) = \ln x$ is continuous, we can interchange the function and the limit on the right-hand side:

$\ln L = \displaystyle \lim_{n \to \infty} \ln \left[ P \left( 1 + \frac{r}{n} \right)^{nt} \right]$

$\ln L = \displaystyle \lim_{n \to \infty} \left[ \ln P + \ln \left( 1 + \frac{r}{n} \right)^{nt} \right]$

$\ln L = \displaystyle \lim_{n \to \infty} \left[ \ln P + nt \ln \left( 1 + \frac{r}{n} \right)\right]$

$\ln L = \ln P + \displaystyle \lim_{n \to \infty} \frac{t \displaystyle \ln \left(1 + \frac{r}{n} \right)}{1/n}$

$\ln L = \ln P + \displaystyle t \lim_{n \to \infty} \frac{\displaystyle \ln \left(1 + \frac{r}{n} \right)}{1/n}$

The limit on the right-hand side follows the indeterminate form $0/0$, as so we may apply L’Hopital’s Rule. Taking the derivative of both the numerator and denominator with respect to $n$, we find

$\ln L = \ln P + \displaystyle t \lim_{n \to \infty} \frac{\displaystyle \frac{1}{1 + \frac{r}{n}} \cdot \frac{-r}{n^2}}{\displaystyle \frac{-1}{n^2}}$

$\ln L = \ln P + \displaystyle t \lim_{n \to \infty}\frac{r}{1 + \frac{r}{n}}$

$\ln L = \ln P + \displaystyle t \frac{r}{1 + 0}$

$\ln L = rt + \ln P$

We now solve for the original limit $L$:

$L = e^{rt + \ln P}$

$L = e^{rt} e^{\ln P}$

$L = Pe^{rt}$

# Different definitions of e (Part 9): Connecting the two definitions

In this series of posts, I consider how two different definitions of the number $e$ are related to each other. The number $e$ is usually introduced at two different places in the mathematics curriculum:

1. Algebra II/Precalculus: If $P$ dollars are invested at interest rate $r$ for $t$ years with continuous compound interest, then the amount of money after $t$ years is $A = Pe^{rt}$.
2. Calculus: The number $e$ is defined to be the number so that the area under the curve $y = 1/x$ from $x = 1$ to $x = e$ is equal to $1$, so that

$\displaystyle \int_1^e \frac{dx}{x} = 1$.

These two definitions appear to be very, very different. One deals with making money. The other deals with the area under a hyperbola. Amazingly, these two definitions are related to each other. In this series of posts, I’ll discuss the connection between the two.

We begin with the second definition, which is usually considered the true definition of $e$. From this definition, I have shown in a previous post that we can derive the differentiation formulas

$\displaystyle \frac{d}{dx} (\ln x) = \frac{1}{x} \qquad$ and $\qquad \displaystyle \frac{d}{dx} \left( e^x \right) = e^x$

beginning with this definition of the number $e$.

Theorem. $\displaystyle \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt} = Pe^{rt}$.

Proof #1.In an earlier post in this series, I showed that

$\displaystyle \frac{1}{x} = \displaystyle \lim_{h \to 0} \ln \left(1 + \frac{h}{x} \right)^{1/h}$

Let’s now replace $h$ with $1/n$. Also, replace $x$ with $1/r$. Then we obtain

$r = \displaystyle \lim_{n \to \infty} \ln \left(1 + \frac{1/n}{1/r} \right)^{n}$

$r = \displaystyle \lim_{n \to \infty} \ln \left(1 + \frac{r}{n} \right)^{n}$

Multiply both sides by $t$:

$rt = \displaystyle t \lim_{n \to \infty} \ln \left(1 + \frac{r}{n} \right)^{n}$

$rt = \displaystyle \lim_{n \to \infty} t \ln \left(1 + \frac{r}{n} \right)^{n}$

$rt = \displaystyle \lim_{n \to \infty} \ln \left[ \left(1 + \frac{r}{n} \right)^{n} \right]^t$

$rt = \displaystyle \lim_{n \to \infty} \ln \left(1 + \frac{r}{n} \right)^{nt}$

Since $g(x) = \ln x$ is continuous, we can interchange the function and the limit on the right-hand side:

$rt = \displaystyle \ln \left[ \lim_{n \to \infty} \left(1 + \frac{r}{n} \right)^{nt} \right]$

$e^{rt} = \displaystyle \lim_{n \to \infty} \left(1 + \frac{r}{n} \right)^{nt}$

Finally, we multiply both sides by $P$:

$P e^{rt} = \displaystyle \lim_{n \to \infty} P \left(1 + \frac{r}{n} \right)^{nt}$

(A second proof of this theorem, using L’Hopital’s Rule, will be presented in tomorrow’s post.)

This firmly established, at last, the connection between the continuous compound interest formula and the area under the hyperbola. I’ve noted that my students feel a certain sense of accomplishment after reaching this point of the exposition.