mathop {lim }limits_{n to infty } left[ {frac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The sum of the cubes of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2 = \frac{n^2

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(C) The sum of the cubes of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} ight]^2 = \frac{n^2(n+1)^2}{4}$.Substituting this into the limit expression:$\mathop {\lim }\limits_{n 	o \infty } \frac{n^2(n+1)^2}{4n^4} = \mathop {\lim }\limits_{n 	o \infty } \frac{n^2(n^2 + 2n + 1)}{4n^4}$.$= \mathop {\lim }\limits_{n 	o \infty } \frac{n^4 + 2n^3 + n^2}{4n^4}$.Divide numerator and denominator by $n^4$:$= \mathop {\lim }\limits_{n 	o \infty } \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{4} = \frac{1 + 0 + 0}{4} = \frac{1}{4}$.

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{1^3} + {2^3} + {3^3} + \dots + {n^3}}}{{{n^4}}}} \right] = $

Similar Questions

The true statement for $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {2 + 3x} - \sqrt {2 - 3x} }}$ is

The value of $\mathop {\lim }\limits_{x \to - 1} \frac{{{x^2} + 3x + 2}}{{{x^2} + 4x + 3}}$ is equal to

If $f: R \rightarrow R$ is defined by $f(x) = [x-3] + |x-4|$ for $x \in R$,then $\lim_{x \rightarrow 3^{-}} f(x)$ is equal to

If $f(x) = \begin{cases} |x|+1, & x < 0 \\ 0, & x = 0 \\ |x|-1, & x > 0 \end{cases}$,for what value$(s)$ of $a$ does $\lim_{x \to a} f(x)$ exist?

$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 1} - x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module