Q104 / 170 · Manipal (MET) 2020 · Mathematics

$\lim _{x \rightarrow 2} \frac{2-\sqrt{2+x}}{2^{1 / 3}-(4-x)^{1 / 3}}$ is equal to

  • A
    $2 \cdot 3^{-1 / 2}$
  • B
    $3 \cdot 2^{-4 / 3}$
  • Correct answer
    C
    $-3 \cdot 2^{-4 / 3}$
  • D
    None of these

Solution

Given, $\lim _{x \rightarrow 2} \frac{2-\sqrt{2+x}}{2^{1 / 3}-(4-x)^{1 / 3}} \quad\left[\frac{0}{0}\right.$ form $]$ $=\lim _{x \rightarrow 2} \frac{0-\frac{1}{2 \sqrt{2+x}}}{0-\frac{1}{3} \cdot \frac{(-1)}{(4-x)^{2 / 3}}}$ $\begin{aligned} & =\lim _{x \rightarrow 2} \frac{-3(4-x)^{2 / 3}}{2 \sqrt{2+x}} \\ & =\frac{-3}{2} \times \frac{(4-2)^{2 / 3}}{\sqrt{2+2}}=\frac{-3}{2} \times \frac{2^{2 / 3}}{2} \\ & =-3 \cdot 2^{\frac{2}{3}-2}=-3 \cdot 2^{-4 / 3}\end{aligned}$

Hence, the correct answer is option: 'C'.

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