What is \lim_{x\rightarrow 1}\frac{f(x)-1}{g(x)} equal to?
Consider the following for the next two (02) items that follow: Let f(x)=\frac{a^{x-1}+b^{x-1}}{2} and g(x)=x-1.
- A. \frac{\ln(ab)}{4}
- B. \frac{\ln(ab)}{2} ✓
- C. \ln(ab)
- D. 2\ln(ab)
Correct Answer: B. \frac{\ln(ab)}{2}
Explanation
Substitute y = x-1, so y \rightarrow 0. The limit becomes \lim_{y \rightarrow 0} \frac{\frac{a^y+b^y}{2}-1}{y} = \lim_{y \rightarrow 0} \frac{a^y+b^y-2}{2y} = \frac{1}{2} \left[\lim_{y \rightarrow 0} \frac{a^y-1}{y} + \lim_{y \rightarrow 0} \frac{b^y-1}{y}\right]. Using the standard limit \lim_{y \rightarrow 0} \frac{k^y-1}{y} = \ln k, we get \frac{1}{2}(\ln a + \ln b) = \frac{\ln(ab)}{2}.
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