_ x 0 ( ₃ 3 x )^ _x 8 =

Step-by-step solution

Consider the limit _ x 0 ( ₃(3x))^ _x 8 . For x 0^+ , the base ₃(3x) = 1 + ₃ x - and the exponent _x 8 = 8 x 0 , yielding the form (- )^0 . Real exponentiation a^b for negative a is undefined when b is irrational, which occurs for x near 0^+ , so the limit does not exist in R . Interpreting the expression as _ x 0^+ | ₃(3x)|^ _x 8 ensures a positive base. With t = ₃ x , t - , and |1 + t| = -(1 + t) for t , we evaluate L = _ t - (-(1+t))^ ₃ 8 / t . Taking natural logarithms: L = ₃ 8 _ t - (-1-t) t . Substitute u = -