I discovered this interesting result while working on another problem.

Can you solve it?

I have the answer; please PM me if you'd like to see it. I'll post it here in a few days. Good luck!

Cheers,

~ Justin

Printable View

- Jul 30th 2013, 10:47 AMjdmcInteresting result of lim k—»0
I discovered this interesting result while working on another problem.

Can you solve it?

I have the answer; please PM me if you'd like to see it. I'll post it here in a few days. Good luck!

Cheers,

~ Justin - Jul 30th 2013, 11:26 AMHallsofIvyRe: Interesting result of lim k—»0
Doesn't look particularly difficult or interesting to me. Just apply L'Hopital's rule twice.

- Jul 30th 2013, 11:27 AMebainesRe: Interesting result of lim k—»0
The limit as k approaches zero is x^2. Take the the second derivative of the numerator and denominator with respect to k, and the limit as k goes to zero is x^2.

- Jul 30th 2013, 01:52 PMjdmcRe: Interesting result of lim k—»0
Well, having forgotten all about L'Hôpital's rule, I thought it was interesting. And I think a student taking calculus might have found it interesting. Oh well,

is right:__ebaines__

I see there's a tough crowd here. My apologies for bringing a butter knife to a gunfight. - Jul 30th 2013, 02:13 PMebainesRe: Interesting result of lim k—»0
- Jul 30th 2013, 02:14 PMHallsofIvyRe: Interesting result of lim k—»0
Ah, come on- yes, the limit is . But please let us see how you got that? Perhaps we were using a cannon when a screwdriver would work!

- Jul 30th 2013, 02:22 PMjdmcRe: Interesting result of lim k—»0
Even though the problem is easy to solve with the use of L'Hôpital's rule, from an intuitive perspective I still find the result interesting.

While testing an approach to another problem, I started with the basic function:

I wanted to modify it so that it would satisfy , and came up with:

Then I wanted to modify it further so that it would intersect , and came up with:

Finally, I wanted to add an argument to vary the curvature of the graph while maintaining the preceding requirements, so I changed it to:

While playing around with different values of in a graphing app, I discovered (graphically, rather than analytically) that the function collimated with when . That was completely unexpected, and, I thought, intriguing… regardless of how elementary the analytic proof is.