Conundrum #3

In calculating the limit $\lim_{x \to 0} f(x)$ we run into the problem that both numerator and denominator tend to zero. Use the L'Hôpital rule.

If $N(x) = 2 - (2^8 - 7x)^{1/8}$ and $D(x) = (5x - 2^5)^{1/5} - 2$ then $N'(x) = 7\cdot \tfrac18 \cdot (2^8 - 7x)^{-7/8};\ \ \ \ \ D'(x) = 5\cdot \tfrac15\cdot (5x - 2^5)^{-4/5}.$ Now $\lim_{x\to 0} \frac{N(x)}{D(x)} = \frac{N'(0)}{D'(0)} = \frac{\tfrac 78(2^8)^{-7/8}}{(2^5)^{-4/5}} = 7\cdot \frac{2^{-3}\cdot 2^{-7}}{2^{-4}} = 7\cdot 2^{-6} = \frac7{64}.$ The answer is therefore "None of these" ( $\boxed{\mathbf D}$ ).