MCQ
$\mathop {Lim}\limits_{n\, \to \,\infty } \,\int\limits_0^2 {{{\left( {1 + \frac{t}{{n + 1}}} \right)}^n}dt} $ is equal to
  • A
    $0$
  • B
    $e^2$
  • $e^2 - 1$
  • D
    does not exist

Answer

Correct option: C.
$e^2 - 1$
c
$\mathop {Lim}\limits_{n\, \to \,\infty } \,\int\limits_0^2 {{{\left( {1 + \frac{t}{{n + 1}}} \right)}^n}dt} $

$=\mathop {Lim}\limits_{n\, \to \,\infty } \,\left[ {{{\left( {1 + \frac{t}{{n + 1}}} \right)}^{n + 1}}} \right]_{\,0}^{\,2}$

 $=\mathop {Lim}\limits_{n\, \to \,\infty } \,{\left( {1 + \frac{2}{{n + 1}}} \right)^{n + 1}} - 1$

 $= e^2 - 1$  $\left[ {\left( {1 + \frac{t}{{n + 1}}} \right){\rm{ is\,  a\,  linear\,  function}}} \right]$

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