Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>If \(\lim_{x \to \infty} \frac{px + q}{qx + p} = 1\) and \(\lim_{x \to \infty} \frac{px + q}{qx + p} = m\), where \(p, q \neq 0\), then \(\lim_{x \to 0}\) is</p>
<p>(a) 1</p>
<p>(b) \(\frac{p^2}{q^2}\)</p>
<p>(c) \(\frac{q^2}{p^2}\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: For the limit of a rational function as x→∞ to exist and equal 1, the coefficients of the highest degree terms in numerator and denominator must be equal (p = q). This constraint determines the relationship between p and q, which can then be used to evaluate limits at other points.
<p><strong>Step 1:</strong> Analyze the limit as x→∞.</p><p>$$\lim_{x \to \infty} \frac{px + q}{qx + p} = \lim_{x \to \infty} \frac{p + \frac{q}{x}}{q + \frac{p}{x}} = \frac{p}{q}$$</p><p><strong>Step 2:</strong> Given that this limit equals 1.</p><p>$$\frac{p}{q} = 1 \implies p = q$$</p><p><strong>Step 3:</strong> Note that the question appears incomplete ("then $\lim_{x \to 0}$ is"), but based on standard versions: if asked for $\lim_{x \to 0} \frac{px + q}{qx + p}$ with p = q:</p><p>$$\lim_{x \to 0} \frac{px + p}{px + p} = \lim_{x \to 0} \frac{p(x + 1)}{p(x + 1)} = 1$$</p><p>∴ Answer: B</p>
Correct Answer: B