Question

Find the value of \[\mathop {\lim }\limits_{n \to \infty } \dfrac{{\left[ {1 + 2 + \ldots .n} \right]}}{{3{n^2} + 5}}\] .
A. \[\dfrac{1}{3}\]
B. \[\dfrac{1}{5}\]
C. \[\dfrac{1}{6}\]
D. 6

Answer 1

Hint: First we simplify the given limit function \[\mathop {\lim }\limits_{n \to \infty } \dfrac{{\left[ {1 + 2 + \ldots .n} \right]}}{{3{n^2} + 5}}\] and use the formula of the sum of \[n\] natural numbers. After that, we simplified the expression by taking common factors from the numerator and denominator. After that, using limiting rule and obtain the final result.

Formula used:
Sum of Natural Numbers, \[S = \dfrac{{n[n + 1]}}{2}\]
Limiting rule \[\mathop {\lim }\limits_{n \to a } f(x) = f(a)\]

Complete Step by step solution
Given: \[\mathop {\lim }\limits_{n \to \infty } \dfrac{{\left[ {1 + 2 + \ldots .n} \right]}}{{3{n^2} + 5}}\]
We know that the sum of \[n\] natural number \[1 + 2 + ... + n = \dfrac{{n(n + 1)}}{2}\]
Substituting in the formula \[S = \dfrac{{n[n + 1]}}{2}\]
\[ = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\dfrac{{n[n + 1]}}{2}}}{{3{n^2} + 5}}\]
Taking out the \[\dfrac{1}{2}\] from the expression,
\[ = \dfrac{1}{2}\mathop {\lim }\limits_{n \to \infty } \dfrac{{n[n + 1]}}{{3{n^2} + 5}}\]
\[ = \dfrac{1}{2}\mathop {\lim }\limits_{n \to \infty } \dfrac{{{n^2} + n}}{{3{n^2} + 5}}\]
Dividing the numerator and denominator by \[{n^2}\], we get
\[ = \dfrac{1}{2}\mathop {\lim }\limits_{n \to \infty } \dfrac{{\dfrac{{{n^2} + n}}{{{n^2}}}}}{{\dfrac{{3{n^2} + 5}}{{{n^2}}}}}\]
Taking the \[n^{2}\] common and cancelling it from both the numerator and denominator
\[ = \dfrac{1}{2}\mathop {\lim }\limits_{n \to \infty } \dfrac{{1 + \dfrac{1}{n}}}{{3 + \dfrac{5}{{{n^2}}}}}\]
As here \[n \to \infty \], so \[\dfrac{1}{n}\] limit tends to zero, we get
Thus,
\[ = \dfrac{1}{2}\left[ {\dfrac{{1 + 0}}{{3 + 0}}} \right]\]
\[ = \dfrac{1}{2}\left[ {\dfrac{1}{3}} \right]\]
\[ = \dfrac{1}{6}\]

Thus, the correct answer is option (C)

Note: The Student should be very diligent when knowing what AP formula to use as it can change the whole answer if not, the student as well should be careful with how to use the limit function and the calculation should be done precisely.