Limits, Continuity & Differentiability
Limit with Greatest Integer Function
Grade 12
Question:
<p>The value of <span>\(\lim_{n \to \infty} \frac{[r] + [2r] + \cdots + [nr]}{n^2}\)</span>, where <span>\(r\)</span> is a non-zero real number and <span>\([x]\)</span> denotes the greatest integer function, is equal to</p>
<p>(A) \(\dfrac{r}{2}\)</p>
<p>(B) \(r\)</p>
<p>(C) \(2r\)</p>
<p>(D) \(0\)</p>
Step-by-Step Solution
Key Concept: Use the inequality property of the greatest integer function and the squeeze theorem to find the limit by bounding the sum.
<p><strong>Step 1:</strong> Use the property of the greatest integer function:</p><p>$r \leq [r] < r + 1$</p><p>$2r \leq [2r] < 2r + 1$</p><p>$nr \leq [nr] < nr + 1$</p><p><strong>Step 2:</strong> Add all inequalities:</p><p>$r + 2r + \cdots + nr \leq [r] + [2r] + \cdots + [nr] < (r + 2r + \cdots + nr) + n$</p><p><strong>Step 3:</strong> Simplify using the sum formula:</p><p>$\frac{n(n+1)}{2}r \leq [r] + [2r] + \cdots + [nr] < \frac{n(n+1)}{2}r + n$</p><p><strong>Step 4:</strong> Divide by $n^2$:</p><p>$\frac{n(n+1)r}{2n^2} \leq \frac{[r] + [2r] + \cdots + [nr]}{n^2} < \frac{n(n+1)r + 2n}{2n^2}$</p><p><strong>Step 5:</strong> Take the limit as $n \to \infty$:</p><p>$\lim_{n \to \infty} \frac{[r] + [2r] + \cdots + [nr]}{n^2} = \frac{r}{2}$</p><p>∴ Answer is (A) $\dfrac{r}{2}$</p>
Correct Answer: A