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

[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'

Homework 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.

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 )

 

[SammyS]

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'

Homework 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.

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 ?

 

[Mark44]

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

 

[AGNuke]

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

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

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.

 

[Fightfish]

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 doesn't guarantee that the limit exists, for one may reach 0 faster than the other.

 

[AGNuke]

it doesn't 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.

 

[Fightfish]

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

 

[dimension10]

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'

Homework 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.

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.

 

[dimension10]

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

Thats like saying the limit doesn't exist.

 

[SammyS]

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.

 

[AGNuke]

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

 

[SammyS]

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

Yes.

How about the denominator?

 

[AGNuke]

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.

 

[SammyS]

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 .

 

[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]

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?

 

[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.

 

[AGNuke]

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.

 

[AGNuke]

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)

 

[SammyS]

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.

 

[klondike]

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.

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)

 

[klondike]

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-:)

 

[AGNuke]

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.