Limits of Sin(n)/n: Solving for N

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In summary, the limit of sin(n)/n as n approaches infinity is equal to 0. Using the definition of limit, it is possible to show that for any small value of epsilon, there exists a value of n greater than a certain number, such that the absolute value of sin(n)/n will be less than epsilon. This can be done by using the fact that sin(anything) is always less than or equal to 1. Therefore, the limit is equal to 0.
  • #1
mjjoga
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lim (sin⁡(n)/n)=0.
The instruction say I can only use the definition of limit and no additional theorems.
So the first thing I should do is figure out if l sin(n)/n l < epsilon, find out what n is greater than. I can pull the 1/n out of the absolute value, but I don't know how to get the sine out. I mean, I know that sin(anything) would be less than say 2, but I don't know if I can do that step with the absolute value still there. Once I can define N>... I can do the rest of it.
Thanks for any help,
mjjoga
 
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  • #2
Are you allowed to use geometric arguments? If you are, that is actually the best approach here.
 
  • #3
welcome to pf!

hi mjjoga! welcome to pf! :smile:
mjjoga said:
lim (sin⁡(n)/n)=0.
…
I know that sin(anything) would be less than say 2

(i assume that's the limit as n -> ∞ ?)

well, sin(anything) is ≤ 1, isn't it? :wink:

yes, if you use that the proof should be easy :smile:
 
  • #4


tiny-tim said:
hi mjjoga! welcome to pf! :smile:


(i assume that's the limit as n -> ∞ ?)

well, sin(anything) is ≤ 1, isn't it? :wink:

yes, if you use that the proof should be easy :smile:

Oh, it's to infinity...

I thought it was to 0. My bad, ignore my post!
 
  • #5
he he :biggrin:
 
  • #6
Yes, it is to infinity, my book wrote it just like that though, it's weird. I was just worried about the abs. value messing it up, but even with it it will be less than or equal to 1. Thanks guys, I appreciate it.
mjjoga
 

FAQ: Limits of Sin(n)/n: Solving for N

What is the limit of sin(n)/n as n approaches infinity?

The limit of sin(n)/n as n approaches infinity is equal to 0. This means that as n gets larger and larger, the value of sin(n)/n gets closer and closer to 0.

How do you solve for n in sin(n)/n = x?

To solve for n in sin(n)/n = x, you can use algebraic manipulations to isolate n on one side of the equation. In this case, you would multiply both sides by n to get sin(n) = nx. Then, you can use trigonometric identities or a calculator to find the value of n that satisfies the equation.

Can the limit of sin(n)/n be negative?

No, the limit of sin(n)/n cannot be negative. This is because sin(n) oscillates between -1 and 1 as n increases, and dividing by n, which continues to get larger, will always result in a value between -1 and 1. Therefore, the limit must be either 0 or undefined.

What is the relationship between the limit of sin(n)/n and the unit circle?

The limit of sin(n)/n is closely related to the unit circle. As n approaches infinity, the values of sin(n)/n approach 0, which means that the graph of sin(n)/n gets closer and closer to the x-axis. This is equivalent to the unit circle getting closer and closer to the x-axis, since the unit circle is defined by the equation x^2 + y^2 = 1, where sin(n)/n can be represented as y/x.

Is there a way to approximate the limit of sin(n)/n for large values of n?

Yes, there are several methods for approximating the limit of sin(n)/n for large values of n. One method is to use the Taylor series expansion for sin(x) and then divide by n, which will give you a polynomial expression for sin(n)/n that can be evaluated for large values of n. Another method is to use the squeeze theorem, which states that if a function is sandwiched between two other functions that have the same limit, then the function in the middle must also have the same limit. In this case, you can use the fact that -1/n ≤ sin(n)/n ≤ 1/n and take the limit as n approaches infinity to find that the limit of sin(n)/n must also be 0.

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