the limit represents the derivative of some function f at some number a. state such an f and a. \n\\( \\lim…

the limit represents the derivative of some function f at some number a. state such an f and a. \n\\( \\lim _ { h \\rightarrow 0 } \\frac { \\sqrt 4 { 16 + h } - 2 } { h } \\)\n\\( \\bigcirc f ( x ) = \\sqrt 4 { x }, a = 2 \\)\n\\( \\bigcirc f ( x ) = \\sqrt { x }, a = 4 \\)\n\\( \\bigcirc f ( x ) = \\sqrt 4 { x }, a = 16 \\)\n\\( \\bigcirc f ( x ) = \\sqrt { x }, a = 16 \\)\n\\( \\bigcirc f ( x ) = x ^ { 4 }, a = 2 \\)

the limit represents the derivative of some function f at some number a. state such an f and a. \n\\( \\lim _ { h \\rightarrow 0 } \\frac { \\sqrt 4 { 16 + h } - 2 } { h } \\)\n\\( \\bigcirc f ( x ) = \\sqrt 4 { x }, a = 2 \\)\n\\( \\bigcirc f ( x ) = \\sqrt { x }, a = 4 \\)\n\\( \\bigcirc f ( x ) = \\sqrt 4 { x }, a = 16 \\)\n\\( \\bigcirc f ( x ) = \\sqrt { x }, a = 16 \\)\n\\( \\bigcirc f ( x ) = x ^ { 4 }, a = 2 \\)

Answer

Explanation:

Step1: Recall the definition of the derivative

The definition of the derivative of a function (y = f(x)) at (x=a) is (f^{\prime}(a)=\lim_{h\rightarrow0}\frac{f(a + h)-f(a)}{h}).

Step2: Compare the given limit with the derivative formula

Given (\lim_{h\rightarrow0}\frac{\sqrt[4]{16 + h}-2}{h}). If we set (f(x)=\sqrt[4]{x}) and (a = 16), then (f(a+h)=\sqrt[4]{16 + h}) and (f(a)=\sqrt[4]{16}=2). So (\lim_{h\rightarrow0}\frac{f(a + h)-f(a)}{h}=\lim_{h\rightarrow0}\frac{\sqrt[4]{16 + h}-2}{h})

Answer:

(f(x)=\sqrt[4]{x},a = 16) (the third option)