# mixedmath

Explorations in math and programming
David Lowry-Duda

This is the fall 2012 Math 90 Introductory Calculus I posthead for David Lowry's TA sections (which should be those in section 1 with Hulse). This is not the main site for the whole course (which can be found at https://sites.google.com/a/brown.edu/fa12-math0090/), but it will contain helpful bits and is a good venue through which you can ask questions.

In particular, the posts that have been put up so far can be found under the Math 90 tag category. If this is your first time visiting, and you are one of my students, please go to the Math 90: Week 1 page and leave a comment.

Here are links to the pages themselves:

And now, the administrative details (the rest can be found on the main course website).

TA Name: David Lowry

email address: djlowry [at] math [dot] brown.edu (although please only use email for private communication - math questions can be asked here, and others can benefit from their openness).

Instructor Name: Thomas Hulse

Info on how to comment

bold, italics, and plain text are allowed in comments. A reasonable subset of markdown is supported, including lists, links, and fenced code blocks. In addition, math can be formatted using $(inline math)$ or $$(your display equation)$$.

Please use plaintext email when commenting. See Plaintext Email and Comments on this site for more. Note also that comments are expected to be open, considerate, and respectful.

Comment via email

1. 2012-11-11 Robin Sifre

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.

Comment 2: On 2012-11-12 davidlowryduda

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.

That is, $lim_{x to -infty} arctan x = - pi/2$.

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?

Comment 3: On 2012-11-12 Robin Sifre

yes, it makes sense. Although I am also a bit confused about why (+)pi/2 doesn't work, as there is an asymptote there, and it looks like the asymptotes are going to both pos. and neg. infinity?

Comment 4: On 2012-11-12 davidlowryduda

Good question! Since $\tan(\theta)$ can be equal at different values of $\theta$, we have to be careful what we mean when we say arctan. Which point do we choose as the inverse? To make sense of this, the domain of each is chosen to be one "repetition" of the function. For arctan, the relevant domain of tan is $(-\pi/2, \pi/2)$.

You're right that tan nears both $\infty$ and $-\infty$ near $\pi/2$, but when we look only at the domain from $(-\pi/2, \pi/2)$, tan nears only $\infty$ near $pi/2$.

Does that make sense?