MCQ
$\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{4}{{x - 1}}} \right)^{3x - 1}} = $
- A${e^{12}}$
- ✓${e^{ - 12}}$
- C${e^4}$
- D${e^3}$
$ = {e^{\mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{ - 4\left( {3 - \frac{1}{x}} \right)}}{{\left( {1 - \frac{1}{x}} \right)}}} \right]}} = {e^{ - 12}}$.
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