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

AGNuke

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



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

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

[tex]\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}[/tex]

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 )

SammyS

Have t\you tried L'Hôpital's rule ?

Mark44

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

AGNuke

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

[tex]\frac{n^2e^{nx}+n^2e^{-nx}+\frac{n^2}{2}cos\frac{nx}{2}-2k}{-sinx - 2tanx.sex^2x}[/tex]

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

[tex]\frac{5n^2}{4}-k=0[/tex]

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

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.

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.

Fightfish

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

dimension10

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

dimension10

Thats like saying the limit doesn't exist.

SammyS

That all looks good.

[itex]\displaystyle \frac{5n^2}{4}-k=0\quad\to\quad k=\frac{5n^2}{4}[/itex]

So plug-in [itex]\displaystyle \frac{5n^2}{4}[/itex] for k :

[itex]\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)}[/itex]

and call on L'Hôpital one more time.

AGNuke

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

SammyS

Yes.

How about the denominator?

AGNuke

Denominator was 0 right from the beginning.

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

SammyS

What's the derivative of [itex]-\sin x(1 +2\sec^3 x)\ ?[/itex]

That derivative is not zero at x=0 .

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.

SammyS

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

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.