Solving L'Hospital's Rule: lim(x→2)(x^2+x-6)/(x-2)

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Homework Help Overview

The discussion revolves around evaluating the limit of the expression (x^2+x-6)/(x-2) as x approaches 2, specifically using L'Hospital's Rule. The subject area is calculus, focusing on limits and the application of L'Hospital's Rule in cases of indeterminate forms.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply L'Hospital's Rule after identifying a 0/0 form. Some participants suggest alternative methods, such as factoring the numerator to simplify the expression before evaluating the limit.

Discussion Status

The discussion includes affirmations of the original poster's approach, with some participants confirming the validity of using L'Hospital's Rule in this context. There is an acknowledgment of the method's appropriateness for the given limit problem.

Contextual Notes

Participants note the importance of confirming the presence of a 0/0 form before applying L'Hospital's Rule, indicating a shared understanding of the conditions under which the rule is applicable.

phillyolly
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Homework Statement



lim(x→2)(x^2+x-6)/(x-2)

Homework Equations





The Attempt at a Solution



lim(x→2)(x^2+x-6)/(x-2)=lim(x→2)(d/dx(x^2+x-6))/(d/dx(x-2))=lim(x→2)(2x+1)/1=lim(x→2)(2x+1)=5

Is this a right solution? I have never done such before, so I am not sure in my answer.
 
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It's fine if you've checked you have a 0/0 form to begin with. You could also factor x^2+x-6 and cancel the denominator to check.
 
yeaa, it is a right way,absolutely.L'Hopital's Rule is for 0/0 and inf/inf.
 
Thank you a lot, guys.
 

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