Limits, Continuity & Differentiability
Limits involving floor functions and exponentials
Grade 12
Question:
<p>The value of $\lim_{x \to \infty} \frac{[1^2(\sin x)^x] + [2^2(\sin x)^x] + \ldots + [n^2(\sin x)^x]}{n^3}$ (where $[\cdot]$ denotes the greatest integer function) is</p>
<p>(a) $-\frac{x}{3} - \frac{\sin x}{3}$</p>
<p>(b) $-\frac{x}{3}(\sin x)$</p>
<p>(c) $\frac{1}{3}$</p>
<p>(d) $0$</p>
Step-by-Step Solution
Key Concept: Recognize that $(\sin x)^x \to 0$ as $x \to \infty$ when $|\sin x| < 1$, making the floor function values approach a constant.
<p>Since $-1 \leq \sin x \leq 1$, we have $(\sin x)^x \to 0$ as $x \to \infty$ for $|\sin x| < 1$. Therefore $[k^2(\sin x)^x] \to -1$ or $0$ for each $k$. Dividing the sum by $n^3$ and taking the limit gives $0$.</p>
Correct Answer: D