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 Question from Neil, a student: The limit of [(1/x)^3 - 1/8]^1/3 all over (x - 1/2) as x approaches 1/2 to positive infinity. How to answer that?

Hi Neil,

This is a strange problem

$\lim_{x \rightarrow \frac12} \frac{[\frac{1}{x}^3 - \frac18]^{1/3}}{x - \frac12}.$

Are you sure that you have it stated properly?

If this is the correct problem then as $x$ approaches $\large \frac12$ the denominator approaches $0$ and the numerator approaches $[2^3 - \large \frac18]^{1/3}$ which is not zero and hence the limit approaches infinity.

Harley

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