The limit does not exist!

It’s October 3rd.

As all lovers of the movie Mean Girls will know, this is the day that Aaron Samuels asked Cady Heron what day it was. This year is particularly special since October 3rd is on a Wednesday. I’m wearing pink to mark the occasion.

(Fun fact: Mean Girls is set in Evanston and Aaron Samuels goes to Northwestern when he graduates from high school!)

To celebrate Mean Girls Day 2018, let’s take ourselves back to the Mathletes state championship, where Cady Heron went head-to-head with Caroline Krafft in a sudden-death final question.

Every time I watch this scene I cringe slightly at the line, “Find the limit of this equation.” Equations don’t have limits—functions do! But we won’t dwell on this any further.

What I would like to do is establish exactly how Cady concludes that the limit does not exist.

The limit in question is the following:

$$\lim_{x \to 0} \frac{\ln(1-x) – \sin x}{1-\cos^2 x}$$

The naïve approach is to substitute $x=0$ into the top and bottom of the expression.

$$\ln(1-0) – \sin 0  = 0-0 = 0 \quad \text{and} \quad 1-\cos^2 0 = 1-1 = 0$$

Oh no! Our limit has an indeterminate form, namely $\frac{0}{0}$.

This means that we must conclude that the limit does not exist, right?

No it doesn’t! For example, $\lim_{x \to 0} \frac{x}{x}$ has an indeterminate form, and yet the limit very much does exist.

Instead, what we need to do is apply one of the many tools from calculus that are at our disposal. The obvious choice is L’Hôpital’s rule, which tells us that if a limit $\lim_{x \to a} \frac{f(x)}{g(x)}$ has an indeterminate form (and the functions $f,g$ are sufficiently well-behaved around $a$), then it is equal to $\lim_{x \to a} \frac{f'(x)}{g'(x)}$, or does not exist if the latter does not exist.

To make our lives simpler, let’s first make the denominator slightly simpler by observing that $1-\cos^2 x = \sin^2 x$.

Now:

  • $\frac{d}{dx} \left[ \ln(1-x) – \sin x \right] = \frac{-1}{1-x} – \cos x$
  • $\frac{d}{dx} \left[ \sin^2 x \right] = 2\sin x \cos x$

So by l’Hôpital’s rule, we have

$$\lim_{x \to 0} \frac{\ln(1-x) – \sin x}{1-\cos^2 x} = \lim_{x \to 0} \frac{\frac{-1}{1-x} – \cos x}{2\sin x\cos x}$$

Now we’re in business.

Substituting $x=0$ into the numerator yields

$$\frac{-1}{1-0} – \cos 0 = -1 – 1 = -2$$

Substituting $x=0$ into the denominator yields

$$2\sin 0 \cos 0 = 2 \times 0 \times 1 = 0$$

And so the limit has the form $\frac{-2}{0}$.

We can therefore conclude, as Cady correctly did, that the limit does not exist!

The North Shore Mathletes are victorious.

Whatever. I’m getting cheese fries.

New blog!

Hello and welcome! This is a continuation of a blog that was formerly a component of my other website, Infinite Descent, whose main purpose was (and is) to serve as a base for my textbook-in-progress, An infinite descent into pure mathematics.

With the end of my PhD and the beginning of my first ‘real job’ as a lecturer in sight—which will allow me to work more closely on the book—now seemed like a good point to split off the content that does not directly relate to the book itself.

As you might have guessed, this is primarily a blog about mathematics. (The .coffee part of the name is one of many new gTLDs introduced in 2014, and I was amazed to discover that this particular domain name wasn’t taken.) With that said, I might at times post about other things. I have no master plan.

I have a vague idea that I would like to accept guest posts on the blog, particularly from like-minded mathematicians, scholars of teaching and learning, and even students. If there is something you think would fit in well with the blog’s content, please get in touch by sending an email to clive at math dot coffee.