Limits, Continuity & Differentiability
Continuity of products with oscillating factors
Grade 12
Question:
<p>If <span>\(f(x) = \begin{cases} (\sin^{-1} x)^2 \cos\frac{1}{x} & , x \neq 0 \\ 0 & , x = 0 \end{cases}\)</span> for <span>\(x \in (-1, 1)\)</span>, then <span>\(f(x)\)</span> is:</p>
<p>(a) continuous nowhere in <span>\(-1 < x < 1\)</span></p>
<p>(b) continuous everywhere in <span>\(-1 < x < 1\)</span></p>
<p>(c) differentiable nowhere in <span>\(-1 < x < 1\)</span></p>
<p>(d) differentiable everywhere in <span>\(-1 < x < 1\)</span></p>
Step-by-Step Solution
Key Concept: A product of a vanishing term (like <span>$(\sin^{-1} x)^2$</span>) with a bounded oscillating function (like <span>$\cos\frac{1}{x}$</span>) is continuous at the point where the first term vanishes.
<p><strong>Analysis:</strong> At <span>$x = 0$</span>: <span>$\lim_{x \to 0} (\sin^{-1} x)^2 \cos\frac{1}{x} = 0$</span> because <span>$(\sin^{-1} x)^2 \to 0$</span> and <span>$\cos\frac{1}{x}$</span> is bounded, so <span>$f(0) = 0$</span> and the limit matches. For <span>$x \neq 0$</span>, the function is a product of continuous functions. Thus <span>$f(x)$</span> is continuous everywhere in <span>$(-1, 1)$</span>.</p><p>∴ Answer is (b).</p>
Correct Answer: B