Using the squeeze theorem to solve a limit

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Homework Help Overview

The discussion revolves around evaluating a limit that results in the indeterminate form 0/0 when direct substitution is applied. The original poster is tasked with using the squeeze theorem to resolve the limit, but expresses uncertainty about how to proceed.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the formulation of the limit and question whether the original statement of the problem is accurate. There is mention of simplifying the expression algebraically as an alternative to the squeeze theorem. One participant suggests demonstrating an inequality to explore the behavior of the limit.

Discussion Status

The conversation is ongoing, with participants clarifying the problem statement and exploring different approaches. There is no explicit consensus on the best method yet, but some guidance has been offered regarding algebraic simplification.

Contextual Notes

Participants are navigating the constraints of the problem, including the requirement to use the squeeze theorem and the presence of an indeterminate form. The format of the forum posts has also caused some confusion regarding the problem's expression.

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Homework Statement


I have a limit, which when substitution is used the indeterminate form 0/0 occurs. I've been asked to solve the limit by using the squeeze theorem. I really have no idea where to take this.


Homework Equations


Lim (x^2-9)(x-3/|x-3|)
x\rightarrow3


The Attempt at a Solution


I've tried so many attempts at this problem, posting all my work would be massive.
 
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LOLItsAJ said:

Homework Statement


I have a limit, which when substitution is used the indeterminate form 0/0 occurs. I've been asked to solve the limit by using the squeeze theorem. I really have no idea where to take this.


Homework Equations


Lim (x^2-9)(x-3/|x-3|)
x\rightarrow3


The Attempt at a Solution


I've tried so many attempts at this problem, posting all my work would be massive.
Are you sure the problem is what you state, which is equivalent to \displaystyle \lim_{x\to3}\left((x^2-9)\left(x-\frac{3}{|x-3|}\right)\right)\,, or is it \displaystyle \lim_{x\to3}\left((x^2-9)\frac{x-3}{|x-3|}\right)\,?
 
It's your second one. I'm a bit confused by the format used on the forum.
 
It appears to me that simplifying the expression using algebra works better than using the squeeze theorem, but ...

Certainly, \displaystyle (x^2-9)\frac{x-3}{|x-3|}\ge0 for x > 0.

Can you show that \displaystyle (x^2-9)\frac{x-3}{|x-3|}&lt; 8|x-3| for 2 < x < 4 ?
 

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