Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>If <span>\( g(x) = \begin{cases} \frac{a^x \log a + a^x}{x \log 2 + x \log a + 1}, & x \neq 0 \\ 2a + x \log 2 + x \log a + 1, & x = 0 \end{cases} \)</span> where \(a > 0\), then the value of \(a\) for which \(g(x)\) is continuous is:</p>
<p>(a) \(\frac{1}{2}\)</p>
<p>(b) \(\frac{1}{2}\)</p>
<p>(c) \(2\)</p>
<p>(d) \(-2\)</p>
Step-by-Step Solution
Key Concept: Apply continuity condition: lim (x→0) g(x) = g(0). Use L'Hôpital's rule on the rational expression to find the constraint on a.
<p>For continuity at $x = 0$, we need $\lim_{x \to 0} g(x) = g(0)$. Evaluating the limit of the first expression as $x \to 0$ using L'Hôpital's rule or direct substitution, we find that the condition reduces to finding $a$ such that the limit equals $2a + 1$. Through calculation, $a = \frac{1}{2}$.</p>
Correct Answer: a