_ n [ ( n ) n^2 + ( en + 1 n + e ) ]^n is equal to - Vidyadip

MCQ

$\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{\sin \left( n \right)}}{{{n^2}}} + \log \left( {\frac{{en + 1}}{{n + e}}} \right)} \right]^n}$ is equal to

A
$e - \frac{1}{e}$
B
${e^{e - \frac{1}{e}}}$
✓
${e^{\frac{1}{e} - e}}$
D
$\frac{1}{e} - e$

✓

Answer

Correct option: C.

${e^{\frac{1}{e} - e}}$

c
The given form is ${\left( 1 \right)^\infty }$ form

$L = \mathop {\lim }\limits_{n \to \infty } n.\left[ {\frac{{\sin n}}{{{n^2}}} + \log \left( {\frac{{en + 1}}{{n + e}}} \right) - 1} \right]$

$ = \mathop {\lim }\limits_{n \to \infty } \frac{{\sin n}}{n} + n\log \left( {\frac{{ne - 1}}{{ne + {e^2}}}} \right)$

$ = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\sin n}}{n} + \frac{{\log \left( {1 + \frac{{ne - 1}}{{ne + {e^2}}} - 1} \right)}}{{\left( {\frac{{ne - 1}}{{ne + {e^2}}} - 1} \right)}} \times n \times \left( {\frac{{ne - 1}}{{ne + {e^2}}} - 1} \right)} \right]$

$ \Rightarrow \frac{{1 - {e^2}}}{e}$

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