Limits, Continuity & Differentiability
Standard Limits and Indeterminate Forms
Grade 12
Question:
<p>Given that <strong>f(x)</strong> is continuous at <strong>x = 0</strong>, and</p><p>\[\lim_{x \to 0^+} \frac{\sin^3(x) \log(1 + 3x)}{x(\tan^{-1} x)^2 (e^{5x} - 1)} = a\]</p><p>Find the value of <strong>a</strong>.</p>
Step-by-Step Solution
Key Concept: Apply standard limits for trigonometric, logarithmic, and exponential functions; manipulate the expression to recognize these standard forms.
<p><strong>Step 1:</strong> Since the denominator approaches 0 as $x \to 0^+$, the numerator must also approach 0 for the limit to be finite.</p><p><strong>Step 2:</strong> Use standard limits:</p><p>$$\lim_{x \to 0} \frac{\sin x}{x} = 1, \quad \lim_{x \to 0} \frac{\log(1+3x)}{3x} = 1, \quad \lim_{x \to 0} \frac{e^{5x}-1}{5x} = 1$$</p><p><strong>Step 3:</strong> Also, $\lim_{x \to 0} \frac{\tan^{-1} x}{x} = 1$</p><p><strong>Step 4:</strong> Rewrite the limit:</p><p>$$a = \lim_{x \to 0^+} \left[\frac{\sin^3(x)}{x^3} \cdot \frac{\log(1+3x)}{3x} \cdot \frac{5x}{e^{5x}-1} \cdot \frac{(\tan^{-1}x)^2}{x^2}\right]$$</p><p>$$= 1^3 \cdot 1 \cdot 1 \cdot \frac{1}{5} = \frac{1}{5}$$</p><p>Wait, recalculating more carefully:</p><p>$$a = \lim_{x \to 0^+} \left[\frac{\sin^3(x)}{x^3} \cdot \frac{\log(1+3x)}{3x} \cdot \frac{5x}{e^{5x}-1} \cdot \frac{x^2}{(\tan^{-1}x)^2} \cdot \frac{3x}{x}\right]$$</p><p>$$= 1 \cdot 1 \cdot 1 \cdot 1 \cdot 3 \cdot \frac{1}{5} = \frac{3}{5}$$</p><p>∴ $a = \frac{3}{5}$</p>
Correct Answer: 3/5