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

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The limit as x approaches -3 of (x^2 + 6x + 9)/(x - 3) exists and is equal to zero. Both the numerator and denominator are continuous at x = -3, and the denominator does not equal zero at that point. Evaluating the expression results in 0/-6, confirming the limit is zero. Therefore, the limit is simply zero. This demonstrates that the limit exists and is finite.
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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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