# My Favorite One-Liners: Part 99

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Today’s quip is a light-hearted one-liner that I’ll use to lighten the mood when in the middle of a complex calculation, like the following limit problem from calculus:

Let $f(x) = 11-4x$. Find $\delta$ so that $|f(x) - 3| < \epsilon$ whenever $|x-2| < \delta$.

The solution of this problem requires isolating $x$ in the above inequality:

$|(11-4x) - 3| < \epsilon$

$|8-4x| < \epsilon$

$-\epsilon < 8 - 4x < \epsilon$

$-8-\epsilon < -4x < -8 + \epsilon$

At this point, the next step is dividing by $-4$. So, I’ll ask my class,

When we divide by $-4$, what happens to the crocodiles?

This usually gets the desired laugh out of the middle-school rule about how the insatiable “crocodiles” of an inequality always point to the larger quantity, leading to the next step:

$2 + \displaystyle \frac{\epsilon}{4} > x > 2 - \displaystyle \frac{\epsilon}{4}$,

so that

$\delta = \min \left( \left[ 2 + \displaystyle \frac{\epsilon}{4} \right] - 2, 2 - \left[2 - \displaystyle \frac{\epsilon}{4} \right] \right) = \displaystyle \frac{\epsilon}{4}$.

Formally completing the proof requires starting with $|x-2| < \displaystyle \frac{\epsilon}{4}$ and ending with $|f(x) - 3| < \epsilon$.

# My Favorite One-Liners: Part 88

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

In the first few weeks of my calculus class, after introducing the definition of a derivative,

$\displaystyle \frac{dy}{dx} = y' = f'(x) = \lim_{h \to 0} \displaystyle \frac{f(x+h) - f(x)}{h}$,

I’ll use the following steps to guide my students to find the derivatives of polynomials.

1. If $f(x) = c$, a constant, then $\displaystyle \frac{d}{dx} (c) = 0$.
2. If $f(x)$ and $g(x)$ are both differentiable, then $(f+g)'(x) = f'(x) + g'(x)$.
3.  If $f(x)$ is differentiable and $c$ is a constant, then $(cf)'(x) = c f'(x)$.
4. If $f(x) = x^n$, where $n$ is a nonnegative integer, then $f'(x) = n x^{n-1}$.
5. If $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ is a polynomial, then $f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + a_1$.

After doing a few examples to help these concepts sink in, I’ll show the following two examples with about 3-4 minutes left in class.

Example 1. Let $A(r) = \pi r^2$. Notice I’ve changed the variable from $x$ to $r$, but that’s OK. Does this remind you of anything? (Students answer: the area of a circle.)

What’s the derivative? Remember, $\pi$ is just a constant. So $A'(r) = \pi \cdot 2r = 2\pi r$.

Does this remind you of anything? (Students answer: Whoa… the circumference of a circle.)

Generally, students start waking up even though it’s near the end of class. I continue:

Example 2. Now let’s try $V(r) = \displaystyle \frac{4}{3} \pi r^3$. Does this remind you of anything? (Students answer: the volume of a sphere.)

What’s the derivative? Again, $\displaystyle \frac{4}{3} \pi$ is just a constant. So $V'(r) = \displaystyle \frac{4}{3} \pi \cdot 3r^2 = 4\pi r^2$.

Does this remind you of anything? (Students answer: Whoa… the surface area of a sphere.)

By now, I’ve really got my students’ attention with this unexpected connection between these formulas from high school geometry. If I’ve timed things right, I’ll say the following with about 30-60 seconds left in class:

Hmmm. That’s interesting. The derivative of the area of a circle is the circumference of the circle, and the derivative of the area of a sphere is the surface area of the sphere. I wonder why this works. Any ideas? (Students: stunned silence.)

This is what’s known as a cliff-hanger, and I’ll give you the answer at the start of class tomorrow. (Students groan, as they really want to know the answer immediately.) Class is dismissed.

If you’d like to see the answer, see my previous post on this topic.

# My Favorite One-Liners: Part 75

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

The $\delta-\epsilon$ definition of a limit is often really hard for students to swallow:

$\forall \epsilon > 0 \exists \delta > 0 \forall x (0 < |x - c| < \delta \Rightarrow |f(x) - L| < \epsilon)$

To make this a little more palatable, I’ll choose a simple specific example, like $\lim_{x \to 2} x^2 = 4$, or

$\forall \epsilon > 0 \exists \delta > 0 \forall x (0 < |x - 2| < \delta \Rightarrow |x^2 - 4| < \epsilon)$

I’ll use one of the famous lines from “Annie Get Your Gun”:

Anything you can do, I can do better.

In other words, no matter how small a $\delta$ they give me, I can find an $\epsilon$ that meets the requirements of this limit.

# 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:

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

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

$= \sqrt{1 + 0}$

$= 1$.