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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
If $f(x) = \left\{ \begin{array}{l}\frac{{{x^2} - 1}}{{x + 1}},\,{\rm{when \,\,}}x \ne - 1\\\,\,\,\,\,\,\,\, - 2,\,{\rm{when\,\, }}x = - 1\end{array} \right.$,  then
  • A
    $\mathop {\lim }\limits_{x \to {{( - 1)}^ - }} f(x) = - 2$
  • B
    $\mathop {\lim }\limits_{x \to {{( - 1)}^ + }} f(x) = - 2$
  • C
    $f(x)$ is continuous at $x = - 1$
  • D
    All the above are correct
✓

Answer

$\mathop {\lim }\limits_{x \to 1 - } f(x) = - 2$ and $f( - 1) = - \,2.$

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