Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 32

$\displaystyle \lim_{h \to 0} \frac{\sqrt[4]{16 + h} - 2}{h}$ represents the derivative of some function $f$ at some number $a$. State such an $f$ and $a$ in this case.

Recall the limit equation for the derivatives

$\lim\limits_{h \to 0} \displaystyle \frac{f(a + h) - f(a)}{h}$

We can see that the left side equation on the numerator is a square root base 4, So $f(x) = \sqrt[4]{x}$ and $a = 16$ as stated in the limit equation for derivatives.

To check if $a = 16$,

$f(a) = \sqrt[4]{a}$

$f(16) = \sqrt[4]{16} = 2$ just like the value of $f(a)$


Therefore,


$f(x) = \sqrt[4]{x}$ and $a = 16$