Evaluate the limit: lim _{n rightarrow infty} | vedclass

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(C) We are given the limit: $\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$ Since $x$ is independent of $k$,we can write: $\lim _{n

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$\frac{x}{3}$

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(C) We are given the limit: $\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$ Since $x$ is independent of $k$,we can write: $\lim _{n \rightarrow \infty} \frac{x}{n^3} \sum_{k=1}^n k^2$ Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ Substituting this into the expression: $\lim _{n \rightarrow \infty} \frac{x}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}$ $= \lim _{n \rightarrow \infty} \frac{x}{6} \cdot \frac{n(n+1)(2n+1)}{n^3}$ $= \frac{x}{6} \lim _{n \rightarrow \infty} \frac{n^3(1 + \frac{1}{n})(2 + \frac{1}{n})}{n^3}$ $= \frac{x}{6} \cdot (1 \cdot 2) = \frac{2x}{6} = \frac{x}{3}$

Evaluate the limit: $\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$

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