Different Way to Solve a Limit of an Indeterminate Equation

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Homework Help Overview

The discussion revolves around evaluating the limit of an indeterminate form, specifically lim_{x\rightarrow-2}\frac{x^3 + 8}{x+2}. The original poster expresses a desire to find alternative methods to synthetic division for solving this limit.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various methods for solving the limit, including L'Hôpital's rule and factoring the numerator as a sum of cubes. The original poster questions the feasibility of factoring in this case and seeks alternative approaches.

Discussion Status

Some participants have provided insights into L'Hôpital's rule and the factoring of the numerator, noting that these methods can lead to the same result. The conversation reflects a mix of understanding and clarification regarding these techniques, with no explicit consensus on a single preferred method.

Contextual Notes

The original poster expresses a dislike for synthetic division and seeks to understand other approaches, indicating a personal constraint in their problem-solving preferences.

chislam
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[tex] lim_{x\rightarrow-2}\frac{x^3 + 8}{x+2}[/tex]

Ok, I have solved this using synthetic division in which the limit was 12, however I was wondering if there was a way to solve this without using synthetic division (seeing as I hate the process of it).

I've solved plenty of other indeterminate equations by factoring out a polynomial and then being able to cancel out so that I can then plug in x, but I wasn't able to do so with this one.

I'm just curious if there are any other ways to solve this specific limit.

Thanks
 
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L'hospital rule (applies only when it is in forms like 0/0, inf/inf)

differentiate both num and den, and then sub in your -2
 
Yeah I started reading up on that rule, but didn't understand it fully until your reply. So the derivative of the numerator is 3x^2 and the denominator is 1 so then just sub in the -2 and you get 12. Got it, thanks a lot for clearing up that rule for me.
 
Not that this is much different than using synthetic division, but the numerator is a sum of two cubes, which factors according to

[tex]a^3 + b^3 = (a + b) \cdot (a^2 - ab + b^2)[/tex] . So, in fact, this numerator can be factored and the (x + 2) term can be cancelled.
 
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