Finding Limit as x-> 0 of sin and cos equation

[Bro]

Homework Statement

lim (-x + sin(sinx))/(x(-1 + cos(sinx)))
x-> 0

Homework Equations

sinθ/θ=1 as θ-> 0
Squeeze Theorem?

The Attempt at a Solution

Well I've tried doing L'Hopital's rule, but to no avail. Each subsequent derivative just makes the equation nastier and more convoluted.
Eg. lim d/dx(-x + sin(sinx))/d/dx(x(-1 + cos(sinx)))=-1 + cosx(cos(sinx))/(-1 + cos(sinx) - xcosx(sin(sinx)))=the next one is even longer and nothing jumps out at me for a possible solution.
I've also attempted to rewrite the equation but again to no avail.
I've also thought about squeeze theorem, but don't know how to apply it here.
Any ideas?

 

[Dick]

Homework Statement

lim (-x + sin(sinx))/(x(-1 + cos(sinx)))
x-> 0

Homework Equations

sinθ/θ=1 as θ-> 0
Squeeze Theorem?

The Attempt at a Solution

Well I've tried doing L'Hopital's rule, but to no avail. Each subsequent derivative just makes the equation nastier and more convoluted.
Eg. lim d/dx(-x + sin(sinx))/d/dx(x(-1 + cos(sinx)))=-1 + cosx(cos(sinx))/(-1 + cos(sinx) - xcosx(sin(sinx)))=the next one is even longer and nothing jumps out at me for a possible solution.
I've also attempted to rewrite the equation but again to no avail.
I've also thought about squeeze theorem, but don't know how to apply it here.
Any ideas?

Keep going till you stop getting 0/0. It will crack at the third derivative.

 

[haruspex]

Eg. lim d/dx(-x + sin(sinx))/d/dx(x(-1 + cos(sinx)))=-1 + cosx(cos(sinx))/(-1 + cos(sinx) - xcosx(sin(sinx)))

As long as you're careful you can simplify some before proceeding.
cos(sin(x)) is going to be very much like cos(x). The discrepancy is of order x⁴.
In the denominator, the leading term of -1 + cos(sinx) is O(x²). You can throw away all higher order terms in the denominator, and as long as the result is not zero the simplification is justified.

 

[Bro]

As long as you're careful you can simplify some before proceeding.
cos(sin(x)) is going to be very much like cos(x). The discrepancy is of order x⁴.
In the denominator, the leading term of -1 + cos(sinx) is O(x²). You can throw away all higher order terms in the denominator, and as long as the result is not zero the simplification is justified.

Ok, but what is O?

 

[haruspex]

Ok, but what is O?

It's "big O" notation. http://en.wikipedia.org/wiki/Big_O_notation. E.g. in the Taylor expansion sin(x) = x - x³/3! + ... we can write sin(x) = x + O(x³). That is, sin(x) is like x plus terms of order x³ and beyond (for small x).
In the present problem x will go to zero. If the numerator can be reduced to Axⁿ+O(xⁿ⁺¹) then as x approaches the limit we can simplify it to Axⁿ. The xⁿ⁺¹ and beyond terms will become irrelevant. Likewise the denominator.

 

[Bro]

It's "big O" notation. http://en.wikipedia.org/wiki/Big_O_notation. E.g. in the Taylor expansion sin(x) = x - x³/3! + ... we can write sin(x) = x + O(x³). That is, sin(x) is like x plus terms of order x³ and beyond (for small x).
In the present problem x will go to zero. If the numerator can be reduced to Axⁿ+O(xⁿ⁺¹) then as x approaches the limit we can simplify it to Axⁿ. The xⁿ⁺¹ and beyond terms will become irrelevant. Likewise the denominator.

Alright perfect. Thank you both for your help!

 

[Curious3141]

Taylor series applied judiciously gives a quick solution.