_ x ( x + 2 x + 1 )^ x + 3 is - Vidyadip

MCQ

$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 2}}{{x + 1}}} \right)^{x + 3}}$ is

A
$1$
✓
$e$
C
${e^2}$
D
${e^3}$

✓

Answer

Correct option: B.

$e$

b
(b) Let $A = \mathop {\lim }\limits_{x \to \infty } \,{\left( {\frac{{x + 2}}{{x + 1}}} \right)^{x + 3}}$

$ = \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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