\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to - \infty } \left( {\frac{{\rm{n}}}{{{{\rm{n}}^2} + {1^2}}} + \frac{{\rm{n}}}{{{{\rm{n}}^2} + {2^2}}} + \frac{{\rm{n}}}{{{{\rm{n}}^2} + {3^2}}} + \ldots + \frac{1}{{5{\rm{n}}}}} \right)\) is equal to

\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to - \infty } \left( {\frac{{

1.	\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to - \infty } \left( {\frac{{\rm{n}}}{{{{\rm{n}}^2} + {1^2}}} + \frac{{\rm{n}}}{{{{\rm{n}}^2} + {2^2}}} + \frac{{\rm{n}}}{{{{\rm{n}}^2} + {3^2}}} + \ldots + \frac{1}{{5{\rm{n}}}}} \right)\) is equal to
A.	tan-1(3)
B.	tan-1(2)
C.	π/4
D.	π/2
Answer» C. π/4

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\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to - \infty } \left( {\frac{{\rm{n}}}{{{{\rm{n}}^2} + {1^2}}} + \frac{{\rm{n}}}{{{{\rm{n}}^2} + {2^2}}} + \frac{{\rm{n}}}{{{{\rm{n}}^2} + {3^2}}} + \ldots + \frac{1}{{5{\rm{n}}}}} \right)\) is equal to

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