Answer
$\dfrac{1}{32}$
Work Step by Step
We have to determine $L=\lim\limits_{x \to 16} \dfrac{\sqrt[4] x-2}{x-16}$.
Use the factorization formula (twice):
$x^2-y^2=(x-y)(x+y)$
$L=\lim\limits_{x \to 16} \dfrac{\sqrt[4] x-2}{(\sqrt x-4)(\sqrt x+4)}=\lim\limits_{x \to 16} \dfrac{\sqrt[4] x-2}{(\sqrt[4] x-2)(\sqrt[4] x+2)(\sqrt x+4)}$
Simplify:
$L=\lim\limits_{x \to 16} \dfrac{1}{(\sqrt[4] x+2)(\sqrt x+4)}$
Determine the limit:
$L=\dfrac{1}{(\sqrt[4] {16}+2)(\sqrt {16}+4)}=\dfrac{1}{(2+2)(4+4)}=\dfrac{1}{32}$
Note: For factoring $x-16$ we could also use the formula:
$x^n-a^n=(x-a)(x^{n-1}+x^{n-2}a+...+xa^{n-2}+a^{n-1})$
