1) the limit as x approaches -2 of x^3 + 2x^2/x^2 - 2x - 8 A) -2/3 B) 3/2 C) -1/8 D)DNE
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Originally Posted by obsmith08 1) the limit as x approaches -2 of x^3 + 2x^2/x^2 - 2x - 8 A) -2/3 B) 3/2 C) -1/8 D)DNE Note that $\displaystyle \frac{x^3+2x^2}{x^2-2x-8}=\frac{x^2\left(x+2\right)}{\left(x-4\right)\left(x+2\right)}$ Can you take it from here?
Originally Posted by obsmith08 1) the limit as x approaches -2 of x^3 + 2x^2/x^2 - 2x - 8 A) -2/3 B) 3/2 C) -1/8 D)DNE $\displaystyle lim_{x \to -2} \frac{x^3+2x^2}{x^2-2x-8} = -\frac{2}{3}$
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