What is the limit for cos (n pi) and sin (n pi)?

In summary, the conversation is about finding the limit of a given expression involving cos(n pi) and sin(n pi) as n approaches infinity or 1. The limit does not exist for n approaching infinity due to the oscillating nature of sin and cos, while for n approaching 1, the answer can be obtained by substituting n=1 into the expression. However, it is important to note that n is typically considered as an integer value in this context.
  • #1
teng125
416
0
may i know how to solve [ n^2 cos(n(pi)) ]/ n^2 + 42??

i have divided it by n^2 and get cos(n pi) / (42/n^2) and i can't solve already.pls help

what is the limit for cos (n pi) and sin (n pi)??
 
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  • #2
Please use latex or make the expressions more clear, using paranthesis since a/b+c isn't the same as a/(b+c), you see?
Also, the limit for what (of course n here) going to what?

I'm guessing you mean the something which looks like

[tex]\mathop {\lim }\limits_{n \to ?} \frac{{n^2 \cos \left( {n\pi } \right)}}{{n^2 + 42}}[/tex]
 
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  • #3
[ n^2 cos(n pi ) ] / (n^2 + 42) for n >1

what is the limit for cos (n pi) and sin (n pi) for n>1 also??
 
  • #4
teng125 said:
what is the limit for cos (n pi) and sin (n pi) for n>1 also??
Again, the limit of those expressions for n going to what? To 0, pi, infinity, ...? You can't say "the limit for n>1"...
 
  • #5
oh...yaya for n to infinity
 
  • #6
teng125 said:
oh...yaya for n to infinity
Ah :smile:

In that case, the limit doesn't exist since sin as well as cos will keep oscilating between -1 and 1.
 
  • #7
Except for sin(n pi) which is identically zero for all n.
 
  • #8
daveb said:
Except for sin(n pi) which is identically zero for all n.
That's only true for integer values of n, we're letting n go to infinity here.
 
  • #9
so,what is the answer for my first question??
 
  • #10
The answer to this?

[tex]\mathop {\lim }\limits_{n \to +\infty} \frac{{n^2 \cos \left( {n\pi } \right)}}{{n^2 + 42}}[/tex]

The limit doesn't exist for the reason I gave above.
 
  • #11
TD said:
That's only true for integer values of n, we're letting n go to infinity here.
True, but I had always been taught that if you have a limit as "n" goes to infinity, then you are talking about integer values for "n". If you want all values, then you use "x" instead of n. Hence, the reason I made the statement.
 
  • #12
is it possible to obtain an answer if n goes to 1 ??
 
  • #13
teng125 said:
is it possible to obtain an answer if n goes to 1 ??
Sure, just fill in n = 1.
 
  • #14
Gah, shake head, look askance. limit as n goes to 1... n is taken to be integer valued, it makes no sense to ask 'as n tends to 1'. See Daveb's very important interjection, TD.
 

FAQ: What is the limit for cos (n pi) and sin (n pi)?

What is the limit for cos (n pi)?

The limit for cos (n pi) is equal to -1 for all odd integer values of n, and equal to 1 for all even integer values of n.

What is the limit for sin (n pi)?

The limit for sin (n pi) is equal to 0 for all integer values of n.

Is there a general formula for finding the limit of cos (n pi) and sin (n pi)?

Yes, the general formula for finding the limit of cos (n pi) and sin (n pi) is given by:

lim (n→∞) cos (n pi) = (-1)^n

lim (n→∞) sin (n pi) = 0

Why do the limits for cos (n pi) and sin (n pi) differ for odd and even values of n?

This is because the values of cos (n pi) and sin (n pi) oscillate between -1 and 1 as n increases. For odd values of n, the limit will be equal to -1, while for even values of n, the limit will be equal to 1.

What are the practical applications of understanding the limits of cos (n pi) and sin (n pi)?

Understanding the limits of cos (n pi) and sin (n pi) is important in fields such as physics, engineering, and mathematics. These concepts are used in wave and vibration analysis, signal processing, and solving differential equations, among others.

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