mathop {Lim}limits_{n to infty } int_0^2 {lef | vedclass

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(C) Let $I_n = \int_0^2 {\left( {1 + \frac{t}{{n + 1}}} \right)^n} dt$. We know that $\frac{d}{dt} {\left( {1 + \frac{t}{{n + 1}}} \right)^{n + 1}} =

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$e^2 - 1$

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(C) Let $I_n = \int_0^2 {\left( {1 + \frac{t}{{n + 1}}} ight)^n} dt$. We know that $\frac{d}{dt} {\left( {1 + \frac{t}{{n + 1}}} ight)^{n + 1}} = (n + 1) {\left( {1 + \frac{t}{{n + 1}}} ight)^n} \cdot \frac{1}{{n + 1}} = {\left( {1 + \frac{t}{{n + 1}}} ight)^n}$. Therefore,the integral is: $I_n = \left[ {\left( {1 + \frac{t}{{n + 1}}} ight)^{n + 1}} ight]_0^2$. Evaluating at the limits: $I_n = {\left( {1 + \frac{2}{{n + 1}}} ight)^{n + 1}} - {\left( {1 + 0} ight)^{n + 1}} = {\left( {1 + \frac{2}{{n + 1}}} ight)^{n + 1}} - 1$. Taking the limit as $n 	o \infty$: $\mathop {Lim}\limits_{n 	o \infty } I_n = \mathop {Lim}\limits_{n 	o \infty } {\left( {1 + \frac{2}{{n + 1}}} ight)^{n + 1}} - 1$. Using the standard limit $\mathop {Lim}\limits_{x 	o \infty } {\left( {1 + \frac{a}{x}} ight)^x} = e^a$: $\mathop {Lim}\limits_{n 	o \infty } I_n = e^2 - 1$.

$\mathop {Lim}\limits_{n \to \infty } \int_0^2 {\left( {1 + \frac{t}{{n + 1}}} \right)^n} dt$ is equal to

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