Let the function f(x) be defined as: f(x) = cases [ ( 4 + x ) ]^ 1 x , & x 0 k, & x = 0 cases If f(x) is continuous at x = 0 , then the value of k is...

Step-by-step solution

For f(x) to be continuous at x = 0 , we must have f(0) = _ x 0 f(x) . k = _ x 0 [ ( 4 + x ) ]^ 1 x This limit is of the form 1^ . Using the standard limit _ x a f(x)^ g(x) = e^ _ x a g(x)(f(x) - 1) , we get: k = e^ _ x 0 1 x [ ( 4 + x ) - 1 ] Using the identity ( 4 + x ) = 1 + x 1 - x , we have: ( 4 + x ) - 1 = 1 + x 1 - x - 1 = 2 x 1 - x Substituting this back into the limit: k = e^ _ x 0 1 x ( 2 x 1 - x ) k = e^ _ x 0 ( x x ) ( 2 1 - x ) Since _ x 0 x x = 1 and _ x 0 x = 0 , we get: k = e^ 1 2 1 - 0 = e^2 Answer: