MCQ
$x \in R,\,$ માટે $\,\,\mathop {\lim }\limits_{x \to \infty } \,{\left( {\frac{{x - 3}}{{x + 2}}} \right)^x} = . . .$
- A$e$
- B${e^{ - 1}}$
- ✓${e^{ - 5}}$
- D${e^5}$
$\left( {\because \,\mathop {\lim }\limits_{x \to \infty } \frac{{ - 5x}}{{x + 2}} = \,\mathop {\lim }\limits_{x \to \infty } \frac{{ - 5}}{{1 + \frac{2}{x}}} = - 5} \right)$.
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