## Limit of 0/0 indeterminant form, n and k for existence of limit

 Quote by SammyS Then that means that the limit is 0, doesn't it?
Yeah, It was obvious from the start, since that relation between n and k was established assuming limit is 0.

 Quote by klondike Are you allowed to use infinitesimal of equivalent? Note that 3(sin(x)-x) and sin(x)-tan(x) are equivalent infinitesimals as x approaches 0. Which makes much less miserable to apply L'Hôpital's rule multiple times.
Hmm... But my problem is not denominator, it's numerator.

What I actually need is to find the value of n so I can find k as well using the relation derived earlier.

For now, I answered the question through glamorous "hit-and-trial" considering it works best if this problem is to be solved under two minutes. Cancelling 4 with n2 with n = 2 and getting k = 5. (And the limit do exist, 0)

Mentor
 Quote by AGNuke Yeah, It was obvious from the start, since that relation between n and k was established assuming limit is 0.
You got that relation between n and k, by making the limit of the numerator be zero, not by making the limit of the overall expression be zero.

There's a BIG difference.

Ok, I see your confusion. Let me try.
0/0 in essence is comparing relative order of infinitesimals. Because both numerator and denominator are infinitesimal as x ->0. The relative order of infinitesimal can be analyzed by doing Taylor expansion at point x=0, and look at the *FIRST* surviving term. Fornatutely, in this problem, the first surviving terms pop up pretty quickly.

If I have done my math right, the Taylor expansion at x=0 would be:

$$\dfrac{(\dfrac{5n^2}{4}-k)x^2+\dfrac{5n^4}{64} x^4 +o(x^4)}{-\dfrac{1}{2}x^3+o(x^3)}$$

In order for it to have a finite limit, the numerator must not be lower order of infinitesimal than the denominator. That means any lower than cubic power terms in the numerator must die so that numerator is higher order of infinitesimal than the denominator. This is how you get the magic n,k relation.

Once the order of both is determined, the actual limit is determined by the ratio of the second (now the first) surviving term in the numerator and the first in the denominator which is not and will never be dependent of n,k as you can see for the numerator.

Hope that helps.

 Quote by AGNuke Hmm... But my problem is not denominator, it's numerator. What I actually need is to find the value of n so I can find k as well using the relation derived earlier. For now, I answered the question through glamorous "hit-and-trial" considering it works best if this problem is to be solved under two minutes. Cancelling 4 with n2 with n = 2 and getting k = 5. (And the limit do exist, 0)

 Quote by AGNuke it works best if this problem is to be solved under two minutes.
BTW, when I learned calculus 101, we were required to memorize first few terms of some frequently used Taylor expansion, i.e. trigonometric, inverse-trig, ln(1+x), e^x etc. My prof indeed expected us to solve this kind of problem in a close book exam in 2~3 minutes-:)

 Quote by klondike BTW, when I learned calculus 101, we were required to memorize first few terms of some frequently used Taylor expansion, i.e. trigonometric, inverse-trig, ln(1+x), e^x etc. My prof indeed expected us to solve this kind of problem in a close book exam in 2~3 minutes-:)
Well, we are also required to remember expansions if we have any hope to steer clear of problems like these in JEE.