MCQ
If $f(x) = \left\{ \begin{array}{l}\frac{{{x^4} - 16}}{{x - 2}},\,\,{\rm{when}}\,\,x \ne 2\\\,\,\,\,\,\,\,\,\,\,\,\,\,16,\,{\rm{when}}\,\,x = 2\end{array} \right.$, then
  • A
    $f(x)$ is continuous at $x = 2$
  • $f(x)$ is discountinous at $x = 2$
  • C
    $\mathop {\lim }\limits_{x \to 2} f(x) = 16$
  • D
    None of these

Answer

Correct option: B.
$f(x)$ is discountinous at $x = 2$
b
(b) $\mathop {\lim }\limits_{x \to 2} \,f(x) = \mathop {\lim }\limits_{x \to 2} \,(x + 2)\,\,({x^2} + 4) = 32,\,\,f(2) = 16.$

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