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

• Bro
In summary: O(x3) in the denominator means that the simplification will only happen for x near zero, but as long as the numerator remains within that range it will be handled.
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?

Bro said:

## 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.

Bro said:
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.

haruspex said:
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?

Bro said:
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.

haruspex said:
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!

Taylor series applied judiciously gives a quick solution.

## 1. What is the limit as x approaches 0 for the equation sin(x)/x?

The limit as x approaches 0 for the equation sin(x)/x is equal to 1. This can be proven using the Squeeze Theorem or by using the definition of the derivative for the sine function.

## 2. How do I find the limit as x approaches 0 for the equation cos(x)/x?

To find the limit as x approaches 0 for the equation cos(x)/x, you can use L'Hopital's Rule or the definition of the derivative for the cosine function. Both methods will yield a limit of 0.

## 3. Can the limit as x approaches 0 for the equation sin(x)/x be evaluated using a graphing calculator?

Yes, a graphing calculator can be used to evaluate the limit as x approaches 0 for the equation sin(x)/x. Simply plug in values of x that approach 0 and observe the corresponding y-values. The limit should be approximately 1.

## 4. Is it possible for the limit as x approaches 0 for the equation cos(x)/x to not exist?

Yes, it is possible for the limit as x approaches 0 for the equation cos(x)/x to not exist. This would occur if the left and right hand limits do not equal each other or if they approach different values. In this case, we say that the limit does not exist.

## 5. What is the significance of finding the limit as x approaches 0 for these equations?

Finding the limit as x approaches 0 for the equations sin(x)/x and cos(x)/x is important in calculus and other areas of mathematics. These limits are fundamental in understanding the derivatives of trigonometric functions and their applications in real-world problems such as motion and waves.

Replies
3
Views
765
Replies
6
Views
1K
Replies
5
Views
1K
Replies
5
Views
2K
Replies
19
Views
2K
Replies
3
Views
2K
Replies
12
Views
1K
Replies
2
Views
2K
Replies
11
Views
2K
Replies
3
Views
1K