The value of mathop {lim }limits_{x to 0} {le | vedclass

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Question

(D) Let $y = \mathop {\lim }\limits_{x 	o 0} {\left( {\frac{{{a^x} + {b^x} + {c^x}}}{3}} ight)^{2/x}}$.Taking the natural logarithm on both sides:$

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$(abc)^{2/3}$

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(D) Let $y = \mathop {\lim }\limits_{x 	o 0} {\left( {\frac{{{a^x} + {b^x} + {c^x}}}{3}} ight)^{2/x}}$.Taking the natural logarithm on both sides:$\ln y = \mathop {\lim }\limits_{x 	o 0} \frac{2}{x} \ln \left( \frac{a^x + b^x + c^x}{3} ight)$.Since this is a $\frac{0}{0}$ form,we apply $L$'$H$ôpital's rule:$\ln y = 2 \mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx} \ln \left( \frac{a^x + b^x + c^x}{3} ight)}{\frac{d}{dx} (x)}$.$\ln y = 2 \mathop {\lim }\limits_{x 	o 0} \frac{3}{a^x + b^x + c^x} \cdot \frac{a^x \ln a + b^x \ln b + c^x \ln c}{3}$.As $x 	o 0$,$a^x, b^x, c^x 	o 1$:$\ln y = 2 \cdot \frac{1}{3} (\ln a + \ln b + \ln c) = \frac{2}{3} \ln(abc) = \ln((abc)^{2/3})$.Therefore,$y = (abc)^{2/3}$.

The value of $\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{a^x} + {b^x} + {c^x}}}{3}} \right)^{2/x}}$ where $(a, b, c > 0)$ is:

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