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This is the function: $$f(x)= \frac{x^3 + 3x^2+2x}{x^2-2x-8} $$ So I tried to find this limit: $$\lim_{x\to -2}\frac{x^3 + 3x^2+2x}{x^2-2x-8}=\left[\frac{-8-12-4}{4+4-8}=\frac{-24}{0}\right]= \lim_{x\to -2}\frac{x(x-1)(x-2)}{(x-4)(x+2)}$$ This isn't a case where numerator and denominator, after factorization, have some elements which can reduce a "problematic" element... I need to get rid of $(x+2)$ which makes the fraction explode but I don't know how.

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$\color{red}+12$. And $x^3+3x^2+2x=x(x\color{red}+2)(x\color{red}+1)$. –  David Mitra Feb 10 '13 at 0:05
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What is that u doing in the denominator? Shouldnt that be a x? –  CBenni Feb 10 '13 at 0:07
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L'Hôpital's rule –  Elements in Space Feb 10 '13 at 0:07
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@CBenni editor displays 'x' but here it displays 'u'. –  kr85 Feb 10 '13 at 0:10
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If the numerator has finite, non-zero limit, and the denominator has a zero limit, then the limit is infinite or doesn't exist. For example $\lim\limits_{x\rightarrow1} {4\over (x-1)^2}=\infty$ and $\lim\limits_{x\rightarrow1} {4\over x-1 }$ does not exist (the limit from the right is $\infty$ and the limit from the left is $-\infty$). –  David Mitra Feb 10 '13 at 0:28

2 Answers 2

up vote 6 down vote accepted

You have factor wrong your function, just do it in the following way $$\lim_{x\to -2}\frac{x^3 + 3x^2+2x}{x^2-2x-8} = \lim_{x \to -2} \frac{x(x+1)(x+2)}{(x-4)(x+2)} = \lim_{x \to -2} \frac{x(x+1)}{(x-4)} = \frac{-2(-2+1)}{-2-4} = -\frac{1}{3}$$

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I assume you meant $$f(x)=\frac{x^3+3x^2+2x}{x^2-2x-8}$$

Then $\lim_{x\to2-}f(x)$ is a indeterminate form or type $\frac{0}{0}$, therefore we apply l'Hopital rule and receive $$\lim_{x\to2-}f(x)=\lim_{x\to2-}\frac{3x^2+6x+2}{2x-2}=\frac{3\cdot4-2\cdot6+2}{-2\cdot2-2}=\frac{2}{-6}=-\frac{1}{3}$$

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