underset{nto infty }{mathop{lim }},frac{{{1}^ | vedclass

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Question

(B) આપેલ લક્ષ: $\mathop {\lim }\limits_{n 	o \infty } \,\frac{{{1^{99}} + {2^{99}} + {3^{99}} + \dots + {n^{99}}}}{{{n^{100}}}}$આને આ રીતે લખી શકાય:

Vedclass · Accepted Answer

$\frac{1}{100}$

Vedclass · Answer

(B) આપેલ લક્ષ: $\mathop {\lim }\limits_{n 	o \infty } \,\frac{{{1^{99}} + {2^{99}} + {3^{99}} + \dots + {n^{99}}}}{{{n^{100}}}}$આને આ રીતે લખી શકાય: $\mathop {\lim }\limits_{n 	o \infty } \,\sum\limits_{r = 1}^n {\,\left( {\frac{{{r^{99}}}}{{{n^{100}}}}} ight)} $$\frac{1}{n}$ ને બહાર લેતા,આપણને મળે છે: $\mathop {\lim }\limits_{n 	o \infty } \,\frac{1}{n}\,\sum\limits_{r = 1}^n {\,{{\left( {\frac{r}{n}} ight)}^{99}}}$નિશ્ચિત સંકલનની વ્યાખ્યા મુજબ,$\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum_{r=1}^n f\left(\frac{r}{n}ight) = \int_0^1 f(x) dx$,જ્યાં $f(x) = x^{99}$.તેથી,પદાવલિ બને છે: $\int_0^1 {{x^{99}}dx}$સંકલનનું મૂલ્ય શોધતા: $\left[ {\frac{{{x^{100}}}}{{100}}} ight]_0^1 = \frac{1^{100}}{100} - \frac{0^{100}}{100} = \frac{1}{100}$.

$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+{{3}^{99}}+......{{n}^{99}}}{{{n}^{100}}}=$

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