Question

If x is an integer with \(x > 1\), the solution of \(\lim_{x \to \infty} \left(\dfrac{1}{x^2} + \dfrac{2}{x^2} + \dfrac{3}{x^2} + \cdots + \dfrac{x-1}{x^2} + \dfrac{1}{x}\right)\) is

• A. Zero
• B. 0.5
• C. 1.0
• D. \(\infty\)

Answer 1

The Correct Option is B

To solve the given problem, we need to evaluate the limit:

\[\lim_{x \to \infty} \left(\dfrac{1}{x^2} + \dfrac{2}{x^2} + \dfrac{3}{x^2} + \cdots + \dfrac{x-1}{x^2} + \dfrac{1}{x}\right)\]

Rewriting the series, we have:

\[\sum_{k=1}^{x-1} \frac{k}{x^2} + \frac{1}{x}\]

The first part of the series can be expressed as:

\[\sum_{k=1}^{x-1} \frac{k}{x^2} = \frac{1}{x^2} \sum_{k=1}^{x-1} k\]

The summation \(\sum_{k=1}^{x-1} k\) represents the sum of the first \(x-1\) natural numbers, which is:

\[\frac{(x-1)x}{2}\]

Therefore, the series becomes:

\[\frac{1}{x^2} \cdot \frac{(x-1)x}{2} + \frac{1}{x}\]

Simplifying further, we get:

\[\frac{(x-1)x}{2x^2} + \frac{1}{x} = \frac{x-1}{2x} + \frac{1}{x}\]

Combining the fractions:

\[\frac{x-1 + 2}{2x} = \frac{x+1}{2x}\]

We now find the limit of this expression as \(x \to \infty\):

\[\lim_{x \to \infty} \frac{x+1}{2x} = \lim_{x \to \infty} \left(\frac{x}{2x} + \frac{1}{2x}\right)\]

Which simplifies to:

\[\lim_{x \to \infty} \left(\frac{1}{2} + \frac{1}{2x}\right) = \frac{1}{2} + 0 = \frac{1}{2}\]

Thus, the solution to the limit is 0.5