Question:
The value of the limit \(\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{1^x} + {2^x} + {3^x} + {4^x}}}{4}} \right)^{1/x}}\) is:
(a) \(1\)
(b) \({3^{1/3!}}\)
(c) \({3!^{1/4}}\)
(d) \({4!^{1/4}}\)

Answer:  \({4!^{1/4}}\)
Solution:

Given that \(x \to 0\), hence \({1^x} \approx {2^x} \approx {3^x} \approx {4^x} \approx 1\)
Since all the numbers are tending to the same value, we can take their Geometric Mean in place of Arithmetic Mean.
\(\left( {\frac{{{1^x} + {2^x} + {3^x} + {4^x}}}{4}} \right) = {\left( {{1^x}{2^x}{3^x}{4^x}} \right)^{1/4}}\)
And the value of the limit = \({\left[ {{{\left( {{1^x}{2^x}{3^x}{4^x}} \right)}^{1/4}}} \right]^{1/x}}\)
\( = {24^{1/4}} = {(4!)^{1/4}}\)