Limits, Continuity & Differentiability
Limits using L'Hôpital or expansion
Grade 12
Question:
<p>Evaluate: \[\lim_{x \to 0} \left( \frac{1}{x^2} \int_0^x \frac{t+t^2}{1+\sin t} \, dt \right)\]</p>
<p>\(\frac{1}{2}\)</p>
<p>\(0\)</p>
<p>\(1\)</p>
<p>\(\frac{1}{3}\)</p>
Step-by-Step Solution
Key Concept: Use L'Hôpital's rule twice on the indeterminate form 0/0, recognizing that the numerator's derivatives involve the integrand evaluated at the limit point. The key is that d/dx[∫₀ˣ f(t)dt] = f(x).
<p><strong>Step 1:</strong> Recognize the indeterminate form. As x→0: numerator→0 and denominator→0, so we have 0/0.</p><p><strong>Step 2:</strong> Apply L'Hôpital's rule (first time):<br/>$$\lim_{x \to 0} \frac{\frac{d}{dx}\left[\int_0^x \frac{t+t^2}{1+\sin t} dt\right]}{\frac{d}{dx}[x^2]} = \lim_{x \to 0} \frac{\frac{x+x^2}{1+\sin x}}{2x}$$</p><p><strong>Step 3:</strong> Simplify:<br/>$$= \lim_{x \to 0} \frac{x+x^2}{2x(1+\sin x)} = \lim_{x \to 0} \frac{x(1+x)}{2x(1+\sin x)} = \lim_{x \to 0} \frac{1+x}{2(1+\sin x)}$$</p><p><strong>Step 4:</strong> Direct substitution is now valid:<br/>$$= \frac{1+0}{2(1+\sin 0)} = \frac{1}{2(1)} = \frac{1}{2}$$</p><p>∴ Answer: <strong>1/2</strong></p>
Correct Answer: A