Solving the Limit of (1-cos x)sin(1/x)

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Homework Help Overview

The problem involves finding the limit of the expression (1 - cos x)sin(1/x) as x approaches 0. The subject area pertains to limits and trigonometric functions.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to reason through the limit by noting that (1 - cos x) approaches 0 and that sin(1/x) oscillates infinitely as x approaches 0. They question how to formally demonstrate that the limit is 0, considering the bounded nature of sin(1/x).

Discussion Status

Some participants suggest using the squeeze theorem as a potential approach, with one participant providing a specific inequality to support this method. There is acknowledgment of a successful application of the squeeze theorem, but no explicit consensus on the overall understanding of the limit has been reached.

Contextual Notes

Participants are discussing the behavior of the functions involved as x approaches 0, specifically focusing on the oscillatory nature of sin(1/x) and the convergence of (1 - cos x) to 0.

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Homework Statement



Find the following limit:

[tex] \lim_{x \to 0} (1-\text{cos }x)\text{sin }\frac{1}{x}[/tex]

Homework Equations





The Attempt at a Solution



(1-cos x) -> 0 as x -> 0. sin (1/x) oscillates infinitely many times as x -> 0.

intuition tells me that the limit is 0, but how do i show that?

some ideas i have are using the fact that |sin(1/x)| =< 1, but I'm not sure.
 
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Try the squeeze theorem with something that converges to zero like [tex]\frac{1-\cos{x}}{x}[/tex].
 
i ended up doing this.

[tex] \begin{align*}<br /> -1 &\leq& \text{sin }\frac{1}{x} &\leq& 1\\<br /> -(1-\text{cos }x) &\leq& (1-\text{cos }x)(\text{sin }\frac{1}{x}) &\leq& 1- \text{cos }x<br /> \end{align*}[/tex]

since both of the terms on the side equal 0 at x=0, by the squeeze theorem, the middle term also goes to 0.
 
That's how I would have done it. Well done!
 

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