mathop {lim }limits_{n to infty } {left[ {fra | vedclass

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(B) माना $P = \mathop {\lim }\limits_{n 	o \infty } {\left( {\frac{{n!}}{{{n^n}}}} ight)^{1/n}}$ है।दोनों पक्षों का प्राकृतिक लघुगणक लेने पर:$\log

Vedclass · Accepted Answer

$1/e$

Vedclass · Answer

(B) माना $P = \mathop {\lim }\limits_{n 	o \infty } {\left( {\frac{{n!}}{{{n^n}}}} ight)^{1/n}}$ है।दोनों पक्षों का प्राकृतिक लघुगणक लेने पर:$\log P = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \log \left( \frac{1 \cdot 2 \cdot 3 \cdot ... \cdot n}{n^n} ight)$$\log P = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum_{r=1}^{n} \log \left( \frac{r}{n} ight)$निश्चित समाकलन की परिभाषा का उपयोग करते हुए,$\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum_{r=1}^{n} f\left( \frac{r}{n} ight) = \int_0^1 f(x) dx$:$\log P = \int_0^1 \log x \, dx$खंडशः समाकलन का उपयोग करते हुए,$\int \log x \, dx = x \log x - x$:$\log P = [x \log x - x]_0^1 = (1 \cdot \log 1 - 1) - \mathop {\lim }\limits_{x 	o 0^+} (x \log x - x)$चूंकि $\mathop {\lim }\limits_{x 	o 0^+} x \log x = 0$,इसलिए:$\log P = -1 - 0 = -1$अतः,$P = e^{-1} = \frac{1}{e}$।

$\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{n!}}{{{n^n}}}} \right]^{1/n}}$ का मान ज्ञात कीजिए।

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