MCQ
$\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{\sin \left( n \right)}}{{{n^2}}} + \log \left( {\frac{{en + 1}}{{n + e}}} \right)} \right]^n}$ =
- A$e - \frac{1}{e}$
- B${e^{e - \frac{1}{e}}}$
- C${e^{\frac{1}{e} - e}}$
- D$\frac{1}{e} - e$
$L = \mathop {\lim }\limits_{n \to \infty } n.\left[ {\frac{{\sin n}}{{{n^2}}} + \log \left( {\frac{{en + 1}}{{n + e}}} \right) - 1} \right]$
$ = \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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સમાંતર શ્રેણીમાં ,હોય તો ${\text{x = }}........$