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

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 )
 PhysOrg.com science news on PhysOrg.com >> Intel's Haswell to extend battery life, set for Taipei launch>> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens

Mentor
 Quote by AGNuke If $\displaystyle \lim_{x \to 0}\frac{e^{nx}+e^{-nx}-2\cos\frac{nx}{2} - kx^2}{\sin x - \tan x}$ exists and finite, then possible values of 'n' and 'k' 2. Relevant equations By far, I have got one equation relating n and k. $\displaystyle \frac{5n^2}{4}-k=0$ 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 indeterminacy in the denominator and expanding the numerator, I got $\displaystyle \frac{\left [2\left ( \frac{(nx)^2}{2!}+\frac{(nx)^4}{4!}+... \right )-2\left ( -\frac{(nx/2)^2}{2!}+\frac{(nx)^4}{4!}+... \right ) \right ]\cos x}{-\left ( \frac{\sin x}{x} \right )\left ( \frac{1-\cos^2x}{x^2} \right )x^3}$ 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 )
Have t\you tried L'Hôpital's rule ?
 Mentor AGNuke, please don't use SIZE tags. I replaced your itex tags with tex tags.

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

 Quote by SammyS Have t\you tried L'Hôpital's rule ?
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

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

 Quote by Mark44 AGNuke, please don't use SIZE tags. I replaced your itex tags with tex tags.
OK and Thanks. I don't have much experience with latex and their size issue.

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

 Quote by Fightfish it doesnt guarantee that the limit exists
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.
 Ah, I get your point. It looks like there are infinitely many solutions though.

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

 Quote by Fightfish Ah, I get your point. It looks like there are infinitely many solutions though.
Thats like saying the limit doesn't exist.

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

Mentor
 Quote by AGNuke If I do so, wouldn't the numerator becomes zero all on its own?
Yes.

 Quote by SammyS Yes. How about the denominator?
Denominator was 0 right from the beginning.

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

Mentor
 Quote by AGNuke Denominator was 0 right from the beginning. Even if I apply LH rule twice, I still won't get any respectable equation.
What's the derivative of $-\sin x(1 +2\sec^3 x)\ ?$

That derivative is not zero at x=0 .
 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.

Mentor
 Quote by 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.
Then that means that the limit is 0, doesn't it?
 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.