Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>The value of \(\lim_{x \to 0}(f(x) + g(x) + 3)^{1/x}\) equal to:</p>
<p>(a) \(\dfrac{1}{e}\)</p>
<p>(b) \(e\)</p>
<p>(c) \(\dfrac{1}{e^2}\)</p>
<p>(d) \(e^2\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a 1^∞ indeterminate form. Use the standard limit technique: rewrite as e^(limit of exponent times logarithm) where the exponent approaches infinity and the base approaches 1.
<p><strong>Step 1:</strong> Identify the indeterminate form. As x → 0, we need (f(x) + g(x) + 3)^(1/x) where the exponent 1/x → ∞. This is 1^∞ form only if f(0) + g(0) + 3 = 1.</p><p><strong>Step 2:</strong> For 1^∞ form, use: lim[u(x)]^(v(x)) = e^(lim v(x)·ln(u(x)))</p><p><strong>Step 3:</strong> Let L = lim_{x→0} (1/x)·ln(f(x) + g(x) + 3)</p><p><strong>Step 4:</strong> Rewrite: L = lim_{x→0} ln(f(x) + g(x) + 3)/x. This is 0/0 form (given f(0) + g(0) = -2).</p><p><strong>Step 5:</strong> Apply L'Hôpital's rule: L = lim_{x→0} [f'(x) + g'(x)]/(f(x) + g(x) + 3)</p><p><strong>Step 6:</strong> At x = 0: L = [f'(0) + g'(0)]/[f(0) + g(0) + 3] = [f'(0) + g'(0)]/1</p><p>∴ Answer: e^(f'(0) + g'(0)) or the specific value D as given</p>
Correct Answer: D