MCQ
If $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x} - \frac{4}{{{x^2}}}} \right)^{2x}} = {e^3},$ then $'a'$ is equal to
- A$2$
- ✓$\frac {3}{2}$
- C$\frac {1}{2}$
- D$\frac {2}{3}$
$\, = e\left[ {\mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{a}{x} - \frac{4}{{{x^2}}} - 1} \right)2x} \right]$
$\, = e\left[ {\mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{a}{x} - \frac{4}{{{x^2}}} - 1} \right)2x} \right]$
$\therefore 2a = 3 \Rightarrow a = 3/2$
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