Q253 / 270 · NDA 2013 (Phase 2) · Mathematics

What is $\lim _{x \rightarrow 2} \frac{2-x}{x^3-8}$ equal to?

  • A
    $\frac{1}{8}$
  • B
    $-\frac{1}{8}$
  • C
    $\frac{1}{12}$
  • Correct answer
    D
    $-\frac{1}{12}$

Solution

$\lim _{x \rightarrow 2} \frac{2-x}{x^3-8}$ This is in the $\frac{0}{0}$ form, we use L'Hospital rule $\lim _{x \rightarrow 2} \frac{-1}{3 x^2}=\frac{-1}{3(2)^2}=-\frac{1}{12}$

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

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