MCQ
$\mathop {\lim }\limits_{x \to 2} \,\,\frac{{{x^3} - 8}}{{{x^2} - 4}} = . .. $
- ✓$3$
- B$\frac{3}{2}$
- C$1$
- D$0$
Applying $L-$ Hospital's rule, we get
$\mathop {\lim }\limits_{x \to 2} \,\,\frac{{3{x^2}}}{{2x}} = \mathop {\lim }\limits_{x \to 2} \,\,\frac{{3 \times 2 \times 2}}{{2 \times 2}} = 3$.
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