The value of the limit of frac{x^3 - 8}{x^2 - | vedclass

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Question

(A) Given limit: $\lim_{x 	o 2} \frac{x^3 - 8}{x^2 - 4}$.Substituting $x = 2$,we get the indeterminate form $\frac{0}{0}$.Using factorization:$\lim_{

Vedclass · Accepted Answer

$3$

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(A) Given limit: $\lim_{x 	o 2} \frac{x^3 - 8}{x^2 - 4}$.Substituting $x = 2$,we get the indeterminate form $\frac{0}{0}$.Using factorization:$\lim_{x 	o 2} \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)}$$= \lim_{x 	o 2} \frac{x^2 + 2x + 4}{x + 2}$$= \frac{2^2 + 2(2) + 4}{2 + 2} = \frac{4 + 4 + 4}{4} = \frac{12}{4} = 3$.Alternatively,applying $L'	ext{Hospital's rule}$:$\lim_{x 	o 2} \frac{d/dx(x^3 - 8)}{d/dx(x^2 - 4)} = \lim_{x 	o 2} \frac{3x^2}{2x} = \lim_{x 	o 2} \frac{3x}{2} = \frac{3(2)}{2} = 3$.

The value of the limit of $\frac{x^3 - 8}{x^2 - 4}$ as $x \to 2$ is

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