Limit Evaluation: Does $\lim_{x\rightarrow 1} \frac{x^3-1}{x^2-1}$ Exist?

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SUMMARY

The limit evaluation of $\lim_{x\rightarrow 1} \frac{x^3-1}{x^2-1}$ exists and can be simplified by factoring. The numerator $x^3-1$ factors into $(x-1)(x^2+x+1)$, while the denominator $x^2-1$ factors into $(x-1)(x+1)$. By canceling the common factor $(x-1)$, the limit can be evaluated as $\lim_{x\rightarrow 1} \frac{x^2+x+1}{x+1}$, which results in a value of 2 when $x$ approaches 1.

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fstam2
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Here is the question,
Evaluate this limit:
[tex]\lim_{x\rightarrow 1} \frac{x^3-1}{x^2-1}[/tex]
since there is no common factor, this limit does not exist, correct?
Maybe I am missing a basic algebra rule for the numerator to find a common factor.
Thank you.
 
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You can factor an (x-1) out of the numerator. You can see that this must be so since x=1 is a root of x^3-1 and a polynomial must factor out into its roots. You can use synthetic division to find what the other factor should be.
 
There is no common factor?? If p(1)= 0 for any polynomial p then (x-1) must be a common factor! x3-1= (x-1)(x2+ x+ 1) and x2- 1= (x-1)(x+ 1).
 

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