Interesting result of lim k—»0

I discovered this interesting result while working on another problem.

$\displaystyle \lim_{k \to 0}\frac{e^{kx}-kx-1}{e^k-k-1}=\ ?$

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

Re: Interesting result of lim k—»0

Doesn't look particularly difficult or interesting to me. Just apply L'Hopital's rule twice.

Re: 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.

Re: 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, __ebaines__ is right:

$\displaystyle \lim_{k \to 0}\frac{e^{kx}-kx-1}{e^k-k-1}=x^2$

I see there's a tough crowd here. My apologies for bringing a butter knife to a gunfight.

Re: Interesting result of lim k—»0

Quote:

Originally Posted by

**jdmc** I see there's a tough crowd here. My apologies for bringing a butter knife to a gunfight.

I hope we don't come across as being mean! You laid out a challenge so our natural inclination is see if we can meet it. I do agree that the limit is interesting in that the result was a bit of a surprise, at least for me, so thanks for bringing it up.

Re: Interesting result of lim k—»0

Ah, come on- yes, the limit is $\displaystyle x^2$. But please let us see how you got that? Perhaps we were using a cannon when a screwdriver would work!

Re: 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:

$\displaystyle f_1(x)=e^x$

I wanted to modify it so that it would satisfy $\displaystyle f(0)=f'(0)=0$, and came up with:

$\displaystyle f_2(x)=e^x-x-1$

Then I wanted to modify it further so that it would intersect $\displaystyle (1,1)$ , and came up with:

$\displaystyle f_3(x)=\frac{e^x-x-1}{e-1-1}$

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

$\displaystyle f_4(x)=\frac{e^{kx}-kx-1}{e^k-k-1}$

While playing around with different values of $\displaystyle k$ in a graphing app, I discovered (graphically, rather than analytically) that the function collimated with $\displaystyle x^2$ when $\displaystyle k\to0$ . That was completely unexpected, and, I thought, intriguing… regardless of how elementary the analytic proof is.