# Math 90: Week 10

We deviated from our regular course of action this week, so we did not have preset examples to do in classes. So instead, I will say a few things, and this can be the new posthead for questions.

We will have a test next week, but I will not have office hours on Monday and I might miss morning recitation. I’ve asked Tom to be ready to step in, so (barring acts of the gods) there will be class next week. I will still be in MRC next week, and I’m planning on having office hours next Wednesday instead. I will be holding additional office hours, in my office, from 11am to 2pm on Wednesday.

Usually we preview the homework in class, but since we didn’t do that this time and I will not be available to help you with the homework before you have to turn it in, I extra encourage you to use this space to ask any questions you have here.

This entry was posted in Brown University, Math 90, Mathematics and tagged , , , , . Bookmark the permalink.

### 4 Responses to Math 90: Week 10

1. Hospitalized Student says:

Hey David,

I don’t understand the basic definition of L’Hopital’s rule. What’s the point of it? I know how to do it because I have seen you and Tom do it, but why do we do it, and how does the definition of it relate to when we are actually using it?

Thanks –

2. mixedmath says:

L’Hopital’s rule provides a way to evaluate limits that we otherwise don’t know how to evaluate. Calculus is essentially the study of limits – derivatives are limits, and let us optimize functions, determine related rates of change, etc. As one continues to study calculus, one looks deeper into limits and their properties. So if you believe that limits are worth studying, than you might accept L’Hopital’s rule as a method allowing you to evaluate and compare more of them.

L’Hopital’s rule says that if you have two differentiable function $f(x)$ and $g(x)$, and you are trying to compute the limit $lim_{x to c} dfrac{f(x)}{g(x)}$, but both the numerator and the denominator go to $infty$ (or $0$), then the limit we want is in fact $lim_{x to c} dfrac{f'(x)}{g'(x)}$. That is to say that in this case, $lim_{x to c} dfrac{f(x)}{g(x)} = lim_{x to c} dfrac{f'(x)}{g'(x)}$.

For example, let’s consider $lim_{x to infty} dfrac{x}{e^x}$. We do not know what this limit is a priori – the top goes to infinity, the bottom goes to infinity. That’s not a number. Does this limit even exist?

It does exist! Since the top and the bottom both go to infinity, we can take the derivative of the top (to get $1$) and the derivative of the bottom (to get $e^x$ again), and conclude that $lim_{x to infty} dfrac{x}{e^x} = lim_{x to infty} dfrac{1}{e^x}$. The big deal is that we do know how to calculate this limit. The top is $1$, and the bottom goes to $infty$. So the overall limit is $0$. I want to emphasize that we did not know this before – so we have gained something.

More complicated limits lend themselves to more complicated methods, and l’Hopital’s rule often provides a method of simplifying the limits.

3. Robin Sifre says:

I have a question about #16 waaaay back from section 3.9 (inverse trig)
lim as x–>(-)infinity of arctan(x)

I got this one wrong on my homework, and am confused as to how the answer is -(pi/3). I thought the answer would have to be somewhere where there is a vertical asymptote, or where cos=0.

• David Lowry says:

Inverse trig is sort of hard to handle, and I understand the confusion, especially if you were led to believe that the answer is $-pi/3$. The correct answer is $– pi/2$, which is when $cos theta = 0$ as you mentioned.

To repeat, $lim_{x to -infty} arctan x = – pi/2$.

Reading your question, it sounds as though you were going along the right paths of reasoning too – looking at the graph of $tan x$, you expect the answer to be at a vertical asymptote, and in particular the asymptote where the “standard portion” of the tangent curve (the one that passes through the origin) goes to $– infty$.

Does that make sense?