Limits, Continuity & Differentiability
Limits of the form 1^infinity
Grade 12
<p>Evaluate: \(\lim_{x \to \infty} \left(\frac{3x^2 + 1}{4x^2 - 1}\right)^{\frac{x^2}{1+x}}\)</p>
Step-by-Step Solution
Key Concept: Rewrite the fraction as 1 plus a small term, then use the standard form (1 + small)^large = e^(small × large). The exponent grows with x, but the fractional part approaches a constant, requiring careful application of logarithmic limits.
<p><strong>Step 1:</strong> Recognize the indeterminate form 1<sup>∞</sup>. Use the logarithmic approach: L = lim(x→∞) (3x²+1)/(4x²-1))^(x²/(1+x))</p><p>ln L = lim(x→∞) [x²/(1+x)] · ln[(3x²+1)/(4x²-1)]</p><p><strong>Step 2:</strong> Simplify the fraction inside the logarithm:</p><p>(3x²+1)/(4x²-1) = (3 - 1/x²)/(4 - 1/x²) = 3/4 · (1 + 1/(3x²))/(1 - 1/(4x²))</p><p>For small terms: ln[(3x²+1)/(4x²-1)] = ln(3/4) + ln[(1 + 1/(3x²))/(1 - 1/(4x²))]</p><p>≈ ln(3/4) + [1/(3x²) + 1/(4x²)] = ln(3/4) + 7/(12x²)</p><p><strong>Step 3:</strong> Substitute back:</p><p>ln L = lim(x→∞) [x²/(1+x)] · [ln(3/4) + 7/(12x²)]</p><p>= lim(x→∞) [x²ln(3/4)/(1+x) + 7/12 · 1/(1+x)]</p><p><strong>Step 4:</strong> Evaluate each term:</p><p>First term: x²ln(3/4)/(1+x) → -∞ (since ln(3/4) < 0 and x²/(1+x) → ∞)</p><p>Second term: 7/(12(1+x)) → 0</p><p>Therefore: ln L = 0, so L = e⁰ = 1</p><p><strong>Correction:</strong> Recalculate: x²/(1+x) ~ x as x→∞. We need: [x²/(1+x)]·ln(3/4) = x·ln(3/4)·(x/(1+x)) → -∞. But reassess: the base→3/4<1, exponent→∞, giving 0.</p><p>∴ <strong>Answer: 0</strong></p>
Correct Answer: 0