Could you explain the solution of this limit?

[ShayanJ]

I know one way is using hoptial's law but that will be a long way.

\lim _{x \rightarrow 0 } {\frac {\cos \left( \sin \left( x \right) \right) -\cos \left( x \right) }{{x}^{4}}}

thanks

 

[micromass]

Maybe you can multiply numerator and denumerator by cos(sin(x))+cos(x)...

 

[Mentallic]

Just going to throw an idea out there, not sure it's necessarily a valid idea though. Since the limit is approaching 0, could we use the approximation that sin(x)\approx x for small x?

Now that I think about it further, it should be valid.

\lim_{x\to a}f(x)=f(\lim_{x\to a}x) which you can apply to \lim_{x\to 0}cos(sin(x))

 

[hunt_mat]

What about a power series expansion of \cos (\sin (x))? There is a good chance that the first two terms will cancel.

 

[PAllen]

Just going to throw an idea out there, not sure it's necessarily a valid idea though. Since the limit is approaching 0, could we use the approximation that sin(x)\approx x for small x?

Now that I think about it further, it should be valid.

\lim_{x\to a}f(x)=f(\lim_{x\to a}x) which you can apply to \lim_{x\to 0}cos(sin(x))

I think this isn't useful. Obviously, cos(sin(x)) approaches cos(x) as x gets small, but x^4 also gets small. The first power of difference is what is critical, and I don't see how this helps you find that.

(If the first effectively power of difference between cos(sin(x)) and cos(x) is < 4, divergence, =4 limit is some number, >4, zero )

 

[PAllen]

What about a power series expansion of \cos (\sin (x))? There is a good chance that the first two terms will cancel.

Taking the deriviatives for a power series boils down to the same work as L'Hopital. You can try compounding the power series of cos(x) and sin(x), but that gets messy too.

 

[PAllen]

I think L'Hopital is not as bad as it looks. You have to be accurate for the first 3 deriviatives (a bit messy), but for the 4th (which really explodes) you can immediately discard terms dominated by sin (because the denominator is then a constant and you've got <mess> - cos in the numerator; only terms all based on cos will contribute, and all cos will be 1 at 0).

 

[Mentallic]

Just wondering, but is my logic here correct?

Just going to throw an idea out there, not sure it's necessarily a valid idea though. Since the limit is approaching 0, could we use the approximation that sin(x)\approx x for small x?

Now that I think about it further, it should be valid.

\lim_{x\to a}f(x)=f(\lim_{x\to a}x) which you can apply to \lim_{x\to 0}cos(sin(x))

So we can let the numerator become cos(x)-cos(x) since we're dealing with x approaching 0 and sinx approximates to x, thus the limit is 0.

 

[Char. Limit]

What about a power series expansion of \cos (\sin (x))? There is a good chance that the first two terms will cancel.

This could work well.

 

[hunt_mat]

Just did a calculation for the power series of [\cos (\sin (x)) and I arrived at the result:
<br /> \cos (\sin (x))=1-\frac{x^{2}}{2}+\frac{5x^{4}}{4!}<br />

 

[PAllen]

Just did a calculation for the power series of [\cos (\sin (x)) and I arrived at the result:
<br /> \cos (\sin (x))=1-\frac{x^{2}}{2}+\frac{5x^{4}}{4!}<br />

If you did this by computing derivatives, you did exactly the same work as applying L'Hopital's rule.

I get a different answer: 1 not 5 for the 4th term coefficient, thus it would cancel against the cosine power series.

I suspect our difference is in the 3d derivative. I got a pair of terms of the form:

-2 <stuff> + <stuff>

If this were +2 <stuff> + <stuff> , then I would get your result. One of us has made a sign mistake. I checked mine twice, but that is no guarantee... I have no access to symbolic software package.

In any case, my point remains that the avoidance of L'Hopital seems unwarranted. All proposals so far boil down to the same work: computing derivatives up to the non-vanishing terms of the 4th derivative.

[Edit: noting the discrepancy, I checked my work a 3d time, now I agree with the 5].

 

[hunt_mat]

I checked this answer with mathematica.

I think that the only way that you can tackle this question via differentiation.

Having said that, you could examine cos(x-x^3/3!+...) via the addition formulae and examine it like that?

 

[PAllen]

So the upshot is, using series, you end up computing the key value of 5 from the 4th derivative and take:

5/24 - 1/24

Using L'Hopital, you compute the same 5, with the same method, and compute:

(5 -1) / 24

Identical work.

Using the addition formula is a nice alternative, first idea that doesn't involve computing non-vanishing terms of the 4th derivative.

 

[ShayanJ]

Ok people cool down I've solved it.just put power series instead of cos(sin(x)) and cos(x)
seperatley and then use L'Hopital.Its much easier than using L'Hopital at first.

 

[vela]

Just wondering, but is my logic here correct?

So we can let the numerator become cos(x)-cos(x) since we're dealing with x approaching 0 and sinx approximates to x, thus the limit is 0.

You have to be a bit more careful keeping track of the terms of various order of x. You could instead do something like this using a sum-to-product trig identity:

\cos(\sin x)-\cos x = -2 \sin\left(\frac{\sin x - x}{2}\right) \sin\left(\frac{\sin x+x}{2}\right)

The argument of the first sine will be of order x³ while the argument of the second sine will be of order x, so the whole thing will be of order x⁴.

 

[PAllen]

Just wondering, but is my logic here correct?



So we can let the numerator become cos(x)-cos(x) since we're dealing with x approaching 0 and sinx approximates to x, thus the limit is 0.

Except that the limit isn't zero. If you put .01 radians into a calculator, you can see numerically that the limit is 1/6.

 

[Mentallic]

You have to be a bit more careful keeping track of the terms of various order of x. You could instead do something like this using a sum-to-product trig identity:

\cos(\sin x)-\cos x = -2 \sin\left(\frac{\sin x - x}{2}\right) \sin\left(\frac{\sin x+x}{2}\right)

The argument of the first sine will be of order x³ while the argument of the second sine will be of order x, so the whole thing will be of order x⁴.

I don't quite understand that, but I'm guessing it has to do with the taylor series expansions which I haven't studied yet.

Except that the limit isn't zero. If you put .01 radians into a calculator, you can see numerically that the limit is 1/6.

Yeah I checked it myself earlier too, and accidentally had my calculator switched to degrees mode so it looked like it was approaching zero

 

[vela]

I don't quite understand that, but I'm guessing it has to do with the taylor series expansions which I haven't studied yet.

Right. You already know the first-order term of sin x. You just need the next non-vanishing term:

\sin x \cong x - \frac{x^3}{3!}

 

[Mentallic]

Right, but I'm not understanding logically why this changes the limit. As x approaches zero, the value of sin(x)\approx x-x^3/3! is still approaching zero and at the same rate as sin(x)\approx x. Maybe because multiple derivatives are needed to be taken to find the answer and the rate of the rate... etc. is different after sufficient derivatives are taken?

 

[vela]

It's because when you subtract x from sin x, you're left with

\sin x-x \cong (x-x^3/3!)-x = -x^3/6

 

[Mentallic]

Yes obviously, but the limit of that as x approaches zero is still zero, so I don't really see how that's answering my question.

 

[vela]

When x is small, sin(x³) behaves as x³, not x. So the numerator will look like kx⁴, where k is some constant, and the limit will be

\lim_{x \to 0} \frac{\cos(\sin x)-\cos x}{x^4} \approx \lim_{x \to 0} \frac{kx^4}{x^4} = k