\(\mathop {\lim }\limits_{n \to \infty } \frac{{{1^{99}} + {2^{99}} + {3^{99}} + \ldots + {n^{99}}}}{{{n^{100}}}}\)

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{1^{99}} + {2^{99}} +

1.	\(\mathop {\lim }\limits_{n \to \infty } \frac{{{1^{99}} + {2^{99}} + {3^{99}} + \ldots + {n^{99}}}}{{{n^{100}}}}\)
A.	99/100
B.	1/100
C.	1/99
D.	1/101
Answer» C. 1/99

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\(\mathop {\lim }\limits_{n \to \infty } \frac{{{1^{99}} + {2^{99}} + {3^{99}} + \ldots + {n^{99}}}}{{{n^{100}}}}\)

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