Question:

$\lim_{x\to0}\dfrac{a^x+b^x+c^x-3}{x}=$

$\ln(abc)$

$\ln(a+b+c)$

$e^{abc}$

$0$

Step-by-Step Solution

Key Concept: L'Hopital or Taylor expand: $(e^{x\ln a}+e^{x\ln b}+e^{x\ln c}-3)/x\to\ln a+\ln b+\ln c$

$=\ln(abc)$.

Correct Answer: 1