Limits, Continuity & Differentiability
Limits at Infinity
Grade 12
Question:
<p>Let \(f(x) = x + \sqrt{x^2 + 2x}\) and \(g(x) = \sqrt{x^2 + 2x} - x\). Which of the following is/are correct?</p>
<p>(a) \(\lim_{x \to \infty} g(x) = 1\)</p>
<p>(b) \(\lim_{x \to \infty} f(x) = 1\)</p>
<p>(c) \(\lim_{x \to -\infty} f(x) = -1\)</p>
<p>(d) \(\lim_{x \to \infty} g(x) = -1\)</p>
Step-by-Step Solution
Key Concept: When evaluating limits of expressions involving square roots, rationalize by multiplying by the conjugate to eliminate indeterminate forms and determine the behavior as x → ±∞.
<p><strong>Step 1: Analyze g(x) as x → ∞</strong></p><p>For g(x) = √(x² + 2x) - x, rationalize by multiplying by the conjugate:</p><p>g(x) = (√(x² + 2x) - x) · (√(x² + 2x) + x)/(√(x² + 2x) + x)</p><p>= (x² + 2x - x²)/(√(x² + 2x) + x) = 2x/(√(x² + 2x) + x)</p><p><strong>Step 2: Find lim[x→∞] g(x)</strong></p><p>Divide numerator and denominator by x (note: x > 0 as x → ∞):</p><p>g(x) = 2x/(√(x² + 2x) + x) = 2/(√(1 + 2/x) + 1)</p><p>As x → ∞: g(x) → 2/(√1 + 1) = 2/2 = 1</p><p>✓ Option (a) is CORRECT: lim[x→∞] g(x) = 1</p><p><strong>Step 3: Find lim[x→∞] f(x)</strong></p><p>For f(x) = x + √(x² + 2x), as x → ∞:</p><p>f(x) = x(1 + √(1 + 2/x)) → ∞</p><p>✗ Option (b) is INCORRECT: lim[x→∞] f(x) ≠ 1</p><p><strong>Step 4: Analyze f(x) as x → -∞</strong></p><p>For x < -2, we have x² + 2x = x(x + 2) > 0, so √(x² + 2x) is defined.</p><p>As x → -∞, let x = -t where t → ∞:</p><p>f(-t) = -t + √(t² - 2t) = -t + |t|√(1 - 2/t)</p><p>Since t > 0: |t| = t, so f(-t) = -t + t√(1 - 2/t)</p><p>= t(√(1 - 2/t) - 1) = t · (-2/t)/(√(1 - 2/t) + 1) [after rationalization]</p><p>= -2/(√(1 - 2/t) + 1) → -2/(1 + 1) = -1</p><p>✓ Option (c) is CORRECT: lim[x→-∞] f(x) = -1</p><p><strong>Step 5: Verify option (d)</strong></p><p>From Step 2, we found lim[x→∞] g(x) = 1, not -1</p><p>✗ Option (d) is INCORRECT: lim[x→∞] g(x) ≠ -1</p><p><strong>∴ Answer: ACD</strong></p>
Correct Answer: ACD