If $a$ and $b$ are chosen randomly by throwing a pair of fair cubical dice, then the probability that $\displaystyle\lim_{x\to0}\left(\frac{a^x+b^x}{2}\right)^{2/x} = 6$ equals
Step-by-Step Solution
Key Concept: Evaluate the limit: $\lim_{x\to0}\left(\frac{a^x+b^x}{2}\right)^{2/x} = e^{\lim_{x\to0}\frac{2}{x}\ln\frac{a^x+b^x}{2}} = e^{\ln(ab)} = ab$. So $ab=6$.
Limit equals $\sqrt{ab}$... actually $= e^{\ln(ab)} = ab = 6$. Favorable: $(1,6),(2,3),(3,2),(6,1)$ → $P=4/36=1/9$.
Correct Answer: (A) $\dfrac{1}{9}$