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Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.

$ \displaystyle \lim_{x\to -2} \frac{x^3 + 8}{x + 2} $

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03:19

Mengsha Yao

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 4

Indeterminate Forms and l'Hospital's Rule

Derivatives

Differentiation

Volume

Missouri State University

Campbell University

University of Michigan - Ann Arbor

Boston College

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So there is a way to solve this without using little tiles role. We can actually factor X cubed plus eight um to be X plus two yes and x squared minus two acts class four or 2 squared. I mean as a result of that it allows us to cancel the X plus twos just leaving us with um This portion right here. That way when we evaluated at -2 we would end up getting 12 is our answer. Now in this case because it's a cubic function, it would actually be easier to use low Patel's rule and then we explain why all we have to do is take the derivative of the numerator and the derivative of the denominator. That would give us one on the denominator and it would give us three X squared on the numerator. Yeah squared. So with that alone we would end up being able to evaluate it at negative two and once again that would give us 12 as the final answer.

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