limx→0xz2−(z−x)238xz−4x23+8xz34 is equal to

limx→0xz2−(z−x)238xz−4x23+8xz34=limx→0x2zx−x23x38z−4x3+8z3x34=limx→0x4/32z−x3x4/38z−4x3+8z34=2z328z34=1223/3.z

limx→0xz2−(z−x)238xz−4x23+8xz34 is equal to

limx→0xz2−(z−x)238xz−4x23+8xz34=limx→0x2zx−x23x38z−4x3+8z3x34=limx→0x4/32z−x3x4/38z−4x3+8z34=2z328z34=1223/3.z

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$\lim_{x \to 0} \frac{x \sqrt[3]{z^{2} - {(z - x)}^{2}}}{{(\sqrt[3]{8 x z - 4 x^{2}} + \sqrt[3]{8 x z})}^{4}}$ is equal to

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$2^{\frac{21}{3}} z$

$\frac{1}{2^{23 / 3} . z}$

$\frac{z}{2^{11 / 3}}$

answer is B.

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Detailed Solution

$\lim_{x \to 0} \frac{x \sqrt[3]{z^{2} - {(z - x)}^{2}}}{{(\sqrt[3]{8 x z - 4 x^{2}} + \sqrt[3]{8 x z})}^{4}} = \lim_{x \to 0} \frac{x \sqrt[3]{2 z x - x^{2}}}{{(\sqrt[3]{x} \sqrt[3]{8 z - 4 x} + \sqrt[3]{8 z} \sqrt[3]{x})}^{4}}$

$= \lim_{x \to 0} \frac{x^{4 / 3} \sqrt[3]{2 z - x}}{x^{4 / 3} {[\sqrt[3]{8 z - 4 x} + \sqrt[3]{8 z}]}^{4}} = \frac{\sqrt[3]{2 z}}{{[2 \sqrt[3]{8 z}]}^{4}} = \frac{1}{2^{23 / 3} . z}$