# Transforming inverse trig graphs

It all started when I tried to get an interesting variation on graphs of inverse trigonometric functions.  Tiring of constant scale changes and translations of inverse trig graphs, I tried $i(x)=x*tan^{-1}x$ , thinking that this product of odd functions leading to an even function would be a nice, but minor, extension for my students.

I reasoned that because the magnitude of arctangent approached $\frac{\pi}{2}$ as $x\rightarrow\infty$, the graph of $i(x)=x*tan^{-1}x$ must approach $y=\frac{\pi}{2}|x|$ .  As shown and to my surprise, $y=i(x)$ seemed to parallel the anticipated absolute value function instead of approaching it.  Hmmmm…..

If this is actually true, then the gap between $i(x)=x*tan^{-1}x$ and $y=\frac{\pi}{2}|x|$ must be constant.  I suspected that this was probably beyond the abilities of my precalculus students, but with my CAS in hand, I (and they) could compute that limit anyway.

Now that was just too pretty to leave alone.  Because the values of x are positive for the limit, this becomes $\displaystyle y=\frac{\pi}{2}|x|-x*tan^{-1}x=\frac{\pi}{2}x-x*tan^{-1}x=x*(\frac{\pi}{2}-tan^{-1}x)$ .

So, four things my students should see here (with guidance, if necessary) are

1. $i(x)=x*tan^{-1}x$ actually approaches $y=\frac{\pi}{2}|x|-1$,
2. the limit can be expressed as a product,
3. each of the terms in the product describes what is happening to the individual terms of the factors of $i(x)$ as x approaches infinity, and
4. (disturbingly) this limit seems to approach $\infty*0$.  A less-obvious recognition  is that as $x\rightarrow\infty$, $\frac{\pi}{2}-tan^{-1}x$  must behave exactly like $\frac{1}{x}$ because its product with x becomes 1,

But what do I do with this for my precalculus students?

NOTE:  As a calculus teacher, I immediately recognized the $\infty*0$ product as a precursor to L’Hopital’s rule.

$\displaystyle\lim_{x\to\infty} [x*(\frac{\pi}{2}-tan^{-1}x)]\rightarrow\infty*0$
$=\displaystyle\lim_{x\to\infty}\frac{\frac{\pi}{2}-tan^{-1}x}{\frac{1}{x}}\rightarrow\frac{0}{0}$
and this form permits L’Hopital’s rule
$=\displaystyle\lim_{x\to\infty}\frac{\frac{-1}{\displaystyle 1+x^2}}{\frac{-1}{\displaystyle x^2}}$
$\displaystyle=\lim_{x\to\infty}\displaystyle\frac{x^2}{1+x^2}=1$

OK, that proves what the graph suggests and the CAS computes.  Rather than leaving students frustrated with a point in a problem that they couldn’t get past (determining the gap between the suspected and actual limits), the CAS kept the problem within reach.  Satisfying enough for some, I suspect, but I’d love suggestions on how to make this particular limit more attainable for students without invoking calculus.  Ideas, anyone?