Question

\( \mathop {\lim }\limits_{x \to 2} {{{3^x} + {3^{3 - x}} - 12} \over {{3^{ - x/2}} - {3^{1 - x}}}} \) is equal to_______.

Hint:

The limit is of the form 0/0 at x = 2, so differentiate numerator and denominator (L'Hospital's rule) and use \(\frac{d}{dx}a^x = a^x\ln a\).

Answer 1

Correct Answer: 36

Let \(N(x)=3^x+3^{3-x}-12\) and \(D(x)=3^{-x/2}-3^{1-x}\).
At x = 2: \(N=9+3-12=0\) and \(D=3^{-1}-3^{-1}=0\), so the form is 0/0.
Apply L'Hospital's rule:
\(N'(x)=\ln 3\,(3^x-3^{3-x})\Rightarrow N'(2)=\ln 3\,(9-3)=6\ln 3\).
\(D'(x)=\ln 3\,\left(-\tfrac12 3^{-x/2}+3^{1-x}\right)\Rightarrow D'(2)=\ln 3\left(-\tfrac16+\tfrac13\right)=\tfrac{\ln 3}{6}\).
Limit = \(\dfrac{6\ln 3}{\ln 3/6}=36\).
Answer: 36