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Evaluate the given limit

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    \[ \lim_{x \to -2} \frac{x^3+8}{x^2-4}. \]


Here we have a limit of a quotient of polynomials. The denominator is 0 as x \to -2, but simplifying we find this zero is removable,

    \begin{align*}  \lim_{x \to -2} \frac{x^3+8}{x^2-4} &= \lim_{x \to -2} \frac{(x+2)(x^2 -2x + 4)}{(x+2)(x-2)} \\  &= \lim_{x \to -2} \frac{x^2 - 2x + 4}{x-2} \\  &= \frac{12}{-4} \\  &= -3. \end{align*}

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