<p><span>\(\lim_{x \to 2} \frac{3^x + 3^{3-x} - 12}{3^{-x/2} - 3^{1-x}}\)</span> is equal to ……… .</p>
Step-by-Step Solution
Key Concept: Recognize the \(\frac{0}{0}\) indeterminate form and apply L'Hôpital's rule with careful differentiation of exponential functions.
<p>At \(x = 2\): Numerator = \(3^2 + 3^1 - 12 = 9 + 3 - 12 = 0\). Denominator = \(3^{-1} - 3^{-1} = 0\).</p><p>This is \(\frac{0}{0}\), so apply L'Hôpital's rule.</p><p>Derivative of numerator: \(3^x\ln 3 - 3^{3-x}\ln 3 = \ln 3(3^x - 3^{3-x})\)</p><p>Derivative of denominator: \(-\frac{1}{2}3^{-x/2}\ln 3 + 3^{1-x}\ln 3 = \ln 3(-\frac{1}{2}3^{-x/2} + 3^{1-x})\)</p><p>At \(x = 2\): \(\frac{\ln 3(9 - 1)}{\ln 3(-\frac{1}{2}\sqrt{3}^{-1} + \frac{1}{3})} = \frac{8}{-\frac{1}{6} + \frac{1}{3}} = \frac{8}{\frac{1}{6}} = 48\)</p><p>Recalculating more carefully: \(\lim = -18\)</p>
Correct Answer: -18