Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>If \(f(x)\) is continuous and \(f\!\left(\dfrac{9}{2}\right) = \dfrac{2}{9}\), then \(\lim_{x \to 0} f\!\left(\dfrac{1 - \cos 3x}{x^2}\right)\) is equal to</p>
<p>\(\dfrac{9}{2}\)</p>
<p>\(\dfrac{2}{9}\)</p>
<p>\(0\)</p>
<p>\(\dfrac{8}{9}\)</p>
Step-by-Step Solution
Key Concept: Since f is continuous, we can evaluate the limit by first finding the limit of the argument, then applying f to that limit value. Recognize that (1 - cos 3x)/x² has a standard limit form.
<p><strong>Step 1:</strong> Find the limit of the argument as x → 0.</p><p>We need to evaluate: $\lim_{x \to 0} \frac{1 - \cos 3x}{x^2}$</p><p><strong>Step 2:</strong> Use the standard limit formula $\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$</p><p>Let θ = 3x, then: $\lim_{x \to 0} \frac{1 - \cos 3x}{x^2} = \lim_{x \to 0} \frac{1 - \cos 3x}{(3x)^2} \cdot 9 = \frac{1}{2} \cdot 9 = \frac{9}{2}$</p><p><strong>Step 3:</strong> Apply continuity of f.</p><p>Since f is continuous and the argument approaches 9/2:</p><p>$\lim_{x \to 0} f\left(\frac{1 - \cos 3x}{x^2}\right) = f\left(\lim_{x \to 0} \frac{1 - \cos 3x}{x^2}\right) = f\left(\frac{9}{2}\right) = \frac{2}{9}$</p><p>∴ Answer: B</p>
Correct Answer: B