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

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(B) We are given the limit: $\mathop {\lim }\limits_{n 	o \infty } \,\frac{{{1^{99}} + {2^{99}} + {3^{99}} + \dots + {n^{99}}}}{{{n^{100}}}}$This can

Vedclass · Accepted Answer

$\frac{1}{100}$

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(B) We are given the limit: $\mathop {\lim }\limits_{n 	o \infty } \,\frac{{{1^{99}} + {2^{99}} + {3^{99}} + \dots + {n^{99}}}}{{{n^{100}}}}$This can be written as: $\mathop {\lim }\limits_{n 	o \infty } \,\sum\limits_{r = 1}^n {\,\left( {\frac{{{r^{99}}}}{{{n^{100}}}}} ight)} $By taking $\frac{1}{n}$ out,we get: $\mathop {\lim }\limits_{n 	o \infty } \,\frac{1}{n}\,\sum\limits_{r = 1}^n {\,{{\left( {\frac{r}{n}} ight)}^{99}}}$Using the definition of a definite integral as the limit of a sum,$\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum_{r=1}^n f\left(\frac{r}{n}ight) = \int_0^1 f(x) dx$,where $f(x) = x^{99}$.Thus,the expression becomes: $\int_0^1 {{x^{99}}dx}$Evaluating the integral: $\left[ {\frac{{{x^{100}}}}{{100}}} ight]_0^1 = \frac{1^{100}}{100} - \frac{0^{100}}{100} = \frac{1}{100}$.

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

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