Lim _ n , , , _0^2 ( 1 + t n + 1 ) ^n dt is equal to - Vidyadip

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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