What is \lim_{x\rightarrow 1}f(x)^{\frac{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. \sqrt{ab} ✓
- B. ab
- C. 2ab
- D. \frac{\sqrt{ab}}{2}
Correct Answer: A. \sqrt{ab}
Explanation
The limit is of the 1^{\infty} form. We use the formula \lim_{x\rightarrow a} f(x)^{h(x)} = e^{\lim_{x\rightarrow a} h(x)(f(x)-1)}. So, \lim_{x\rightarrow 1} f(x)^{\frac{1}{g(x)}} = e^{\lim_{x\rightarrow 1} \frac{f(x)-1}{g(x)}}. From the previous question, the exponent evaluates to \frac{\ln(ab)}{2} = \ln(\sqrt{ab}). Thus, the limit is e^{\ln(\sqrt{ab})} = \sqrt{ab}.
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