How can the Squeeze/Sandwich theorem be used to solve this limit problem?

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Homework Help Overview

The discussion revolves around applying the Squeeze/Sandwich theorem to evaluate the limit of the expression as x approaches 1, specifically limx→1 [(x-1)2sin(1/(1-x))]. The participants are exploring the behavior of the sine function and its implications for the limit.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the bounds of the sine function and how they relate to the limit. One participant attempts to split the limit into separate parts but expresses confusion about the application of the Squeeze theorem. Another participant suggests using inequalities to establish bounds for the limit.

Discussion Status

There is an ongoing exploration of the limit's behavior, with some participants questioning the nature of the sine function as x approaches 1. While one participant claims to have solved the problem, there is still uncertainty regarding the limit of sin(1/(1-x)) as x approaches 1, indicating a lack of consensus on the interpretation of the oscillatory behavior.

Contextual Notes

Participants are grappling with the implications of the sine function's oscillation and its effect on the limit, particularly as it approaches infinity. There is also mention of the need to consider the limits of the outer expressions in relation to the Squeeze theorem.

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Homework Statement


Use the Squeeze/Sandwich theorem to show the

[tex]limx->1 [(x-1)^{2}sin(\frac{1}{1-x})]=0[/tex]


Thats the lim as x approaches 1 for the whole square bracket

The Attempt at a Solution


i split it up into the two separate limits, but I'm kinda lost on how the sandwhich theorem works, my professor said something along the lines of sin1/1-x has to be between 1 and -1 because all sin are, than he change that to an x and found the limit as x approached zero and the two limits worked out, but this limit has x approaching 1 so I'm stuck.
 
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If we let L be the form you are taking the limit of, then -(x-1)^2<=L<=+(x-1)^2, since the sin is always between -1 and +1, right? What are the outside limits?
 
ok i solved it, it's 0... but i kinda just put what the data should be to obtain zero, how do u find the limit as x approaces 1 for sin(1/(1-x)), how can u tell what the sin of infinity is?
 
It oscillates between -1 and 1. It doesn't have a limit. L oscillates too, but the amplitude of the oscillations get smaller and smaller as it approaches infinity.
 

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