Lim x approaches -3 of (x^2+6x+9)/(x-3)

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Discussion Overview

The discussion revolves around evaluating the limit of the expression (x^2+6x+9)/(x-3) as x approaches -3. Participants explore whether the limit exists and what its value might be, focusing on the continuity of the function and the behavior of the numerator and denominator at that point.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant asks for the limit as x approaches -3 of the given expression and requests an explanation if it does not exist.
  • Another participant notes that both the numerator and denominator are continuous at x = -3 and that the denominator does not equal zero at that point, prompting a discussion about the limit's value.
  • A participant calculates the limit and arrives at a value of 0/-6, questioning if the limit would simply be zero.
  • A later reply affirms that if the expression evaluates to a finite value, then that is the value of the limit.

Areas of Agreement / Disagreement

Participants generally agree that the limit can be evaluated and that it approaches a finite value, but there is some uncertainty regarding the interpretation of the calculations and the final value.

Contextual Notes

There is a lack of detailed exploration regarding the steps taken to evaluate the limit, and the discussion does not clarify any assumptions made about the continuity or behavior of the function near x = -3.

Who May Find This Useful

This discussion may be useful for students or individuals interested in limit evaluation, particularly in the context of continuous functions and rational expressions.

kendalgenevieve
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Determine the limit, if it exists. If not, explain why it does not exist.

lim x approaches -3 of (x^2+6x+9)/(x-3)
 
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kendalgenevieve said:
Determine the limit, if it exists. If not, explain why it does not exist.

lim x approaches -3 of (x^2+6x+9)/(x-3)

The top and bottom are both continuous at x = -3 and the denominator isn't 0 there, so what do you think the limit is?
 
Prove It said:
The top and bottom are both continuous at x = -3 and the denominator isn't 0 there, so what do you think the limit is?

When I solved it I got a 0/-6 so would the limit be just zero?
 
kendalgenevieve said:
When I solved it I got a 0/-6 so would the limit be just zero?

Yes, when the expression evaluates to a finite value, then that is the value of the limit. :)
 

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