Limit of sin(x)/x: Converting to sin(1/x)

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Homework Help Overview

The discussion revolves around finding the limit of the function sin(x)/x as x approaches infinity, specifically relating it to the limit of sin(1/x) as x approaches zero. The original poster expresses uncertainty about converting sin(x) to sin(1/x) in this context.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of the Squeeze Theorem and question the validity of dividing through by x in the context of trigonometric functions. There is also a consideration of whether similar reasoning applies to cos(x).

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the limit and the implications of using the Squeeze Theorem. Some guidance has been offered regarding the application of the theorem, but there is no explicit consensus on the conclusions drawn.

Contextual Notes

Participants are navigating the constraints of the problem, particularly in relation to the behavior of trigonometric functions at infinity and the assumptions involved in switching between sine and cosine functions.

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Homework Statement




Find the following Limit in terms of the number
[tex]\alpha=\lim_{x\rightarrow 0}\frac{sinx}{x}[/tex]

(i)[tex]\lim_{x\rightarrow\infty}\frac{sinx}{x}[/tex]

The Attempt at a Solution



[tex]\alpha=\lim_{x\rightarrow\infty}\frac{sin(1/x)}{1/x}[/tex]
But I don't know how to convert sin(x) to sin(1/x):confused:
[
 
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Hurkyl?
 
I'm not terribly sure how you got

[tex]\alpha=\lim_{x\rightarrow\infty}\frac{sin(1/x)}{1/x}[/tex]

Note that you cannot simply divide through by x when dealing with trigonometric functions. In this case you'll want to use the Squeeze Theorem.

Note that [tex]|sin x| \leq 1[/tex]

This implies that [tex]\left| \frac{sin x}{x} \right| \leq \frac{1}{x}[/tex]

Now you can apply the Squeeze Theorem
 
But from sqeeze theorem we get 0.
 
azatkgz said:
But from sqeeze theorem we get 0.

Yeah, and what's the problem?
 
Ok ,then.
 
Can we in the similar manner say that [tex]\lim_{x\rightarrow\infty}\frac{cosx}{x}=0[/tex]
 
Is there any point in the argument which doesn't hold as a result of switching [tex]sin(x)[/tex] to [tex]cos(x)[/tex]?
 

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