MCQ
$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 2}}{{x + 1}}} \right)^{x + 3}} = . . .$
- A$1$
- ✓$e$
- C${e^2}$
- D${e^3}$
$ = \mathop {\lim }\limits_{x \to \infty } \,{\left( {1 + \frac{1}{{x + 1}}} \right)^{x + 3}}$
$ = \mathop {\lim }\limits_{x \to \infty } \,{\left[ {{{\left( {1 + \frac{1}{{x + 1}}} \right)}^{x + 1}}} \right]^{\frac{{\,(x + 3)}}{{(x + 1)}}}} = e$
$\left\{ \because \,\,\underset{x\to \infty }{\mathop{\lim }}\,\,{{\left( 1+\frac{1}{x+1} \right)}^{x+1}}=e \right.$
and $\mathop {\lim }\limits_{x \to \infty } \frac{{\,(x + 3)}}{{(x + 1)}} = \left. {\mathop {\lim }\limits_{x \to \infty } \frac{{\,\left\{ {1 + (3/x)} \right\}}}{{\left\{ {1 + (1/x)} \right\}}} = 1} \right\}$.
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