



 
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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