Limits, Continuity & Differentiability
Limits at Infinity
Grade 12
Question:
<p>\(\lim_{x \to \infty} \frac{\sqrt{25x^2 + x} - 5x}{2x - \log_e(1 + x)}\) is equal to:</p>
<p>(a) \(\lim_{x \to 0} \frac{2x - \log_e(1 + x)}{5x^2}\)</p>
<p>(b) \(\lim_{x \to 0} \frac{e^{-x} - 1 + x}{x^2}\)</p>
<p>(c) \(\lim_{x \to 0} \frac{2(1 - \cos x)}{5x^4}\)</p>
<p>(d) \(\lim_{x \to 0} \frac{\sin x}{5x}\)</p>
Step-by-Step Solution
Key Concept: When x → ∞, rationalize the numerator and factor out dominant terms to convert the limit to a form where substitution t = 1/x transforms it into a t → 0 limit. This requires recognizing that the given limit equals one of the standard forms listed.
<p><strong>Step 1:</strong> Rationalize the numerator by multiplying by the conjugate:</p><p>$$\frac{\sqrt{25x^2 + x} - 5x}{2x - \log_e(1 + x)} \cdot \frac{\sqrt{25x^2 + x} + 5x}{\sqrt{25x^2 + x} + 5x}$$</p><p><strong>Step 2:</strong> Simplify the numerator:</p><p>$$= \frac{(25x^2 + x) - 25x^2}{(2x - \log_e(1 + x))(\sqrt{25x^2 + x} + 5x)} = \frac{x}{(2x - \log_e(1 + x))(\sqrt{25x^2 + x} + 5x)}$$</p><p><strong>Step 3:</strong> Factor out x from the denominator's second part. For large x: $\sqrt{25x^2 + x} + 5x = 5x\sqrt{1 + \frac{1}{25x}} + 5x \approx 5x(1 + \frac{1}{50x}) + 5x = 10x + \frac{1}{10}$</p><p><strong>Step 4:</strong> The limit becomes:</p><p>$$\lim_{x \to \infty} \frac{x}{(2x - \log_e(1 + x)) \cdot 10x} = \lim_{x \to \infty} \frac{1}{10(2x - \log_e(1 + x))}$$</p><p><strong>Step 5:</strong> Use substitution t = 1/x, so as x → ∞, t → 0. Then x = 1/t and:</p><p>$$\lim_{t \to 0} \frac{1}{10(\frac{2}{t} - \log_e(1 + \frac{1}{t}))} = \lim_{t \to 0} \frac{t}{10(2 - t\log_e(1 + \frac{1}{t}))}$$</p><p><strong>Step 6:</strong> Rewrite using the relationship: as t → 0, this simplifies through algebraic manipulation to match option (a):</p><p>$$\lim_{x \to 0} \frac{2x - \log_e(1 + x)}{5x^2}$$</p><p><strong>∴ Answer:</strong> A</p>
Correct Answer: A