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:
$$\cos (\sin (x))=1-\frac{x^{2}}{2}+\frac{5x^{4}}{4!}$$

 

[PAllen]

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

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

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