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

1. Jun 21, 2012

### AGNuke

If $$\lim_{x \to 0}\frac{e^{nx}+e^{-nx}-2cos\frac{nx}{2} - kx^2}{sinx - tanx}$$ exists and finite, then possible values of 'n' and 'k'

2. Relevant equations
By far, I have got one equation relating n and k.
$$\frac{5n^2}{4}-k=0$$

I can simply put choices in there and get answer, but it would be better if I can get answer from the question itself.

3. The attempt at a solution
By removing the indeterminancy in the denominator and expanding the numerator, I got

$$\frac{\left [2\left ( \frac{(nx)^2}{2!}+\frac{(nx)^4}{4!}+... \right )-2\left ( -\frac{(nx/2)^2}{2!}+\frac{(nx)^4}{4!}+... \right ) \right ]cosx}{-\left ( \frac{sinx}{x} \right )\left ( \frac{1-cos^2x}{x^2} \right )x^3}$$

I expanded the series till x3 or x4. I just want an equation which can confirm the value of n, which is probably 2 (as I am getting a fancy answer, k = 5 )

Last edited by a moderator: Jun 21, 2012
2. Jun 21, 2012

### SammyS

Staff Emeritus
Have t\you tried L'Hôpital's rule ?

3. Jun 21, 2012

### Staff: Mentor

AGNuke, please don't use SIZE tags. I replaced your itex tags with tex tags.

4. Jun 21, 2012

### AGNuke

After applying the L'Hôpital's rule twice, I got

$$\frac{n^2e^{nx}+n^2e^{-nx}+\frac{n^2}{2}cos\frac{nx}{2}-2k}{-sinx - 2tanx.sex^2x}$$

After entering x = 0; since the denominator approaches zero, so must the numerator. After doing so, I got that same previous equation

$$\frac{5n^2}{4}-k=0$$

OK and Thanks. I don't have much experience with latex and their size issue.

5. Jun 21, 2012

### Fightfish

Isn't that still an indeterminate form? Even if both the denominator and numerator approach 0 as x tends to zero, it doesnt guarantee that the limit exists, for one may reach 0 faster than the other.

6. Jun 22, 2012

### AGNuke

But the question states that the limit DO exist. That's why I made both of them to approach 0 as the denominator is going to do that for sure.

And I tried using n = 2 and eventually k = 5, the limit is 0.

7. Jun 22, 2012

### Fightfish

Ah, I get your point. It looks like there are infinitely many solutions though.

8. Jun 22, 2012

### dimension10

Assuming you haven't solved this yet, L' Hospitals's rule looks REALLY tempting here.

9. Jun 22, 2012

### dimension10

Thats like saying the limit doesn't exist.

10. Jun 22, 2012

### SammyS

Staff Emeritus
That all looks good.

$\displaystyle \frac{5n^2}{4}-k=0\quad\to\quad k=\frac{5n^2}{4}$

So plug-in $\displaystyle \frac{5n^2}{4}$ for k :

$\displaystyle \frac{n^2e^{nx}+n^2e^{-nx}+\frac{n^2}{2}\cos\frac{nx}{2}-2k}{-\sin x -2\tan x\sec^2 x} \quad\to\quad \frac{n^2e^{nx}+n^2e^{-nx}+\frac{n^2}{2}\cos\frac{nx}{2}-\frac{5n^2}{2}}{-\sin x(1 +2\sec^3 x)}$

and call on L'Hôpital one more time.

11. Jun 22, 2012

### AGNuke

If I do so, wouldn't the numerator becomes zero all on its own?

12. Jun 22, 2012

### SammyS

Staff Emeritus
Yes.

13. Jun 22, 2012

### AGNuke

Denominator was 0 right from the beginning.

Even if I apply LH rule twice, I still won't get any respectable equation.

14. Jun 22, 2012

### SammyS

Staff Emeritus
What's the derivative of $-\sin x(1 +2\sec^3 x)\ ?$

That derivative is not zero at x=0 .

15. Jun 22, 2012

### AGNuke

If I apply LH rule once after that fancy equation, maybe the denominator is -3, but the numerator becomes 0 all on its own, without leaving a gap for determining the value of n.

16. Jun 22, 2012

### SammyS

Staff Emeritus
Then that means that the limit is 0, doesn't it?

17. Jun 22, 2012

### 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.

18. Jun 23, 2012

### AGNuke

Yeah, It was obvious from the start, since that relation between n and k was established assuming limit is 0.

19. Jun 23, 2012

### 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)

20. Jun 23, 2012

### SammyS

Staff Emeritus
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.