Find the Limit of (1/(x-2) - 12/(x^3-8)) as x Approaches 2 - Homework Help

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Homework Help Overview

The problem involves finding the limit of the expression (1/(x-2) - 12/(x^3-8)) as x approaches 2, which falls under the topic of limits in calculus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the method of combining fractions rather than cross multiplying. There is mention of factoring the difference of cubes and concerns about applying L'Hôpital's Rule.

Discussion Status

The discussion includes attempts to simplify the limit expression and explore different methods of approaching the problem. Some participants provide guidance on combining fractions and factoring, while others express uncertainty about the appropriateness of certain techniques.

Contextual Notes

One participant notes that they have not learned L'Hôpital's Rule or Taylor series, which may limit their approach to the problem.

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Homework Statement


find the limit of:

[tex]\lim_{x\rightarrow2}\frac{1}{x-2}-\frac{12}{x^{3}-8}[/tex]




The Attempt at a Solution



Any help guys? Should i cross multiply then divide and see what i get or is there something else.(besides l'hospital's rule and taylor series because i haven't learned them yet.)
 
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I can't imagine why you would think "cross multiply" here. Of course, what you should do is the subtraction indicated in the problem"
[tex]\frac{1}{x- 2}- \frac{12}{x^3- 8}= \frac{?}{x^3- 8}[/tex]
 
mtayab1994 said:

Homework Statement


find the limit of:

[tex]\lim_{x\rightarrow2}\frac{1}{x-2}-\frac{12}{x^{3}-8}[/tex]




The Attempt at a Solution



Any help guys? Should i cross multiply then divide and see what i get or is there something else.
Cross multiply? You do that when you have an equation that involves two fractions.

Combine the two fractions, and then go from there. Note that the difference of cubes can be factored: a3 - b3 = (a - b)(a2 + ab + b2).
mtayab1994 said:
(besides l'hospital's rule and taylor series because i haven't learned them yet.)
L'Hopital's Rule cannot be applied to this problem because what you have is a difference, not a quotient.
 
ok i got:

[tex]\lim_{x\rightarrow2}\frac{x+4}{x^{2}+2x+4}=\frac{1}{2}[/tex]

is that correct?
 
Thanx.
 

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