Limits, Continuity & Differentiability
Continuity and limits
Grade 12
Question:
<p>Let \(f(x) = \begin{cases} \left(\dfrac{2^x + 3^x + 5^x}{3}\right)^{3/x}, & x \neq 0 \\ k, & x = 0 \end{cases}\). If \(f(x)\) is continuous then the value of \(k\) is equal to:</p>
<p>(a) 10</p>
<p>(b) 15</p>
<p>(c) 20</p>
<p>(d) 30</p>
Step-by-Step Solution
Key Concept: For continuity at x=0, we need lim(x→0) f(x) = f(0) = k. Use logarithmic transformation and L'Hôpital's rule on the exponent to convert the indeterminate form (1^∞) into a solvable limit.
<p><strong>Step 1:</strong> For continuity at x=0, we need lim(x→0) f(x) = k.</p><p><strong>Step 2:</strong> Let y = ((2^x + 3^x + 5^x)/3)^(3/x). Taking ln of both sides:<br/>ln y = (3/x) · ln((2^x + 3^x + 5^x)/3)</p><p><strong>Step 3:</strong> Rewrite as: ln y = (3/x) · [ln(2^x + 3^x + 5^x) - ln 3]<br/>= (3/x) · ln((2^x + 3^x + 5^x)/3)</p><p><strong>Step 4:</strong> As x→0, using Taylor series: 2^x ≈ 1 + x·ln2, 3^x ≈ 1 + x·ln3, 5^x ≈ 1 + x·ln5<br/>So 2^x + 3^x + 5^x ≈ 3 + x(ln2 + ln3 + ln5) = 3 + x·ln30</p><p><strong>Step 5:</strong> Therefore: (2^x + 3^x + 5^x)/3 ≈ 1 + (x·ln30)/3<br/>ln((2^x + 3^x + 5^x)/3) ≈ (x·ln30)/3</p><p><strong>Step 6:</strong> Thus: ln y ≈ (3/x) · (x·ln30)/3 = ln30<br/>So y → e^(ln30) = 30</p><p><strong>Step 7:</strong> Since f must be continuous at x=0: k = 30</p><p>∴ Answer: D</p>
Correct Answer: D