Limits, Continuity & Differentiability
Limits of Powers
Grade 12
Question:
<p>If <span class='math'>\lim_{n \to \infty} (\sin^{-1} x)^{n+1} = 1</span>, find the interval in which <span class='math'>x</span> lies.</p>
<p>(a) <span class='math'>(-\sin 1, \sin 1)</span></p>
<p>(b) <span class='math'>(-1, 1)</span></p>
<p>(c) <span class='math'>(0, 1)</span></p>
<p>(d) <span class='math'>(-1, 0)</span></p>
Step-by-Step Solution
Key Concept: For a limit of the form <span class='math'>\lim_{n \to \infty} a^n</span> to equal 1, we need <span class='math'>|a| = 1</span> and <span class='math'>a = 1</span>. Here, the base must satisfy constraints on the inverse sine function.
<p><strong>Step 1:</strong> For <span class='math'>\lim_{n \to \infty} (\sin^{-1} x)^{n+1} = 1</span> to hold, the base <span class='math'>\sin^{-1} x</span> must satisfy:</p><p><span class='math'>|\sin^{-1} x| < 1</span> (so the limit goes to a finite non-zero value)</p><p><strong>Step 2:</strong> For the limit to equal 1:</p><p><span class='math'>\sin^{-1} x = 1</span></p><p><strong>Step 3:</strong> This is not possible since <span class='math'>\sin^{-1} x \in [-\frac{\pi}{2}, \frac{\pi}{2}]</span> and we need <span class='math'>-1 < \sin^{-1} x < 1</span>.</p><p><strong>Step 4:</strong> The condition requires:</p><p><span class='math'>-1 < \sin^{-1} x < 1</span></p><p><span class='math'>\Rightarrow x \in (-\sin 1, \sin 1)</span></p><p>∴ The answer is (a).</p>
Correct Answer: A