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
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}}

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?

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