Question

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

Explanation:

Step1: Recall the definition of the derivative

The definition of the derivative of a function \(y = f(x)\) at \(x=a\) is \(f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}\).

Step2: Compare the given limit with the derivative formula

Given \(\lim_{h
ightarrow0}\frac{\sqrt[4]{16 + h}-2}{h}\).
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 \(\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}=\lim_{h
ightarrow0}\frac{\sqrt[4]{16 + h}-2}{h}\)

Answer:

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