Quick question about the range of a sine function

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Homework Help Overview

The original poster is investigating the range of the function sin(n^3-2n)/n as n approaches infinity, particularly in the context of a proof about limits.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the behavior of the sine function and its range, with some noting that sin(n) is bounded between -1 and 1. There is uncertainty about how to manipulate the range when the argument of the sine function changes. Questions arise regarding the implications of evaluating the limit as n approaches infinity.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the range of the sine function, but there is no explicit consensus on the final approach to take.

Contextual Notes

There is a focus on the limit as n approaches infinity, and some participants mention the need to consider the function's behavior at n=0 for continuity.

dancergirlie
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Homework Statement


I am doing a proof about limits, and I need to know the range that sin(n^3-2n)/n is within for all n.


Homework Equations





The Attempt at a Solution



I know that sin(n) is in between -1 and 1 for all n, so sin(n)/n would be in between -1/n and 1/n for all n. However, I don't know how to manipulate the range if the inside of the sin(n) function is changed. Any help would be great!
 
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Is this for n->infinity?
if so
sin(n^3-2n) is between -1 and 1
is what you need
then consider 1/n
 
yes this is for n as n approaches infinity. Would it be between -(n^2-2)/(n^3-2n) and (n^2-2)/(n^3-2n)?
 
nevermind i misread your comment, i see that it is in between -1/n and 1/n. Thanks for the help!
 
I would find the limit as n->0 and define the function at n=0 to make it continuous. Then find its local maxes and mins from its derivative.
 

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