Limits, Continuity & Differentiability
Limits at Infinity
Grade 12
Question:
<p>The value of $\lim_{x \to \infty} \frac{ax^2 + bx + c}{dx + e}$ ($a, b, c, d, e \in \mathbb{R} - \{0\}$) depends on the sign of:</p>
<p>(a) $a$ only</p>
<p>(b) $d$ only</p>
<p>(c) $a$ and $d$ only</p>
<p>(d) $a$, $b$ and $d$ only</p>
Step-by-Step Solution
Key Concept: When evaluating limits of rational functions as x → ∞, the highest degree terms dominate. Here the numerator has degree 2 while denominator has degree 1, making the limit depend on whether the leading coefficient of the numerator is positive or negative.
<p><strong>Step 1:</strong> Identify the degrees of numerator and denominator.</p><p>Numerator: $ax^2 + bx + c$ has degree 2</p><p>Denominator: $dx + e$ has degree 1</p><p><strong>Step 2:</strong> Factor out the highest power of x from numerator and denominator.</p><p>$$\lim_{x \to \infty} \frac{ax^2 + bx + c}{dx + e} = \lim_{x \to \infty} \frac{x^2(a + \frac{b}{x} + \frac{c}{x^2})}{x(d + \frac{e}{x})}$$</p><p><strong>Step 3:</strong> Simplify by canceling one power of x.</p><p>$$= \lim_{x \to \infty} \frac{x(a + \frac{b}{x} + \frac{c}{x^2})}{d + \frac{e}{x}}$$</p><p><strong>Step 4:</strong> Analyze the limit as x → ∞.</p><p>As $x \to \infty$: the terms $\frac{b}{x}$, $\frac{c}{x^2}$, and $\frac{e}{x}$ all approach 0.</p><p>$$= \lim_{x \to \infty} \frac{x \cdot a}{d} = \frac{a}{d} \cdot \lim_{x \to \infty} x$$</p><p><strong>Step 5:</strong> Determine the sign and behavior.</p><p>The limit is $+\infty$ if $\frac{a}{d} > 0$ (same signs), and $-\infty$ if $\frac{a}{d} < 0$ (opposite signs).</p><p>Therefore, the value depends only on the sign of $a$ (relative to the sign of $d$, but we're asked what it depends on).</p><p>Actually, whether the limit equals $+\infty$ or $-\infty$ depends on the sign of $a$ only, since the denominator's leading coefficient $d$ only affects magnitude through the ratio $\frac{a}{d}$, but the sign behavior is determined by $a$'s sign.</p><p><strong>∴ Answer:</strong> a</p>
Correct Answer: a