Limits, Continuity & Differentiability
Evaluation of Limits using Properties of Integers
Grade 12
Question:
<p><strong>Ex. 37:</strong> Statement I: $\lim_{m,n \to \infty} \sin(2\pi n! \cdot 3^n) = 0$ when $x$ is rational.</p><p>Statement II: When $n \to \infty$ and $x$ is rational $x = \frac{p}{q}$ where $p, q$ are integers and $q \neq 0$, then $n! \cdot x = n! \cdot \frac{p}{q}$ is an integer.</p>
<p>(a) Statement I is true, Statement II is true; Statement II is correct explanation for Statement I</p>
<p>(b) Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I</p>
<p>(c) Statement I is true, Statement II is false</p>
<p>(d) Statement I is false, Statement II is true</p>
Step-by-Step Solution
Key Concept: For rational $x = \frac{p}{q}$, factorials eventually contain the denominator, making $n! \cdot x$ an integer; $\sin(2\pi k) = 0$ for any integer $k$.
<p><strong>Solution:</strong> When $n \to \infty$ and $x$ is rational, $x = \frac{p}{q}$ where $p, q$ are integers and $q \neq 0$. For $n \geq q$, $n!$ contains $q$ as a factor, so $n! \cdot x = n! \cdot \frac{p}{q}$ is an integer. When $n! \cdot 3^n$ is an integer, $\sin(2\pi n! \cdot 3^n) = 0$ because sine of an integer multiple of $2\pi$ is zero. Therefore, Statement II correctly explains Statement I.</p><p>∴ Answer is (a): Statement I is true, Statement II is true; Statement II is correct explanation for Statement I.</p>
Correct Answer: A