Answer
$\lim\limits_{x\to16}\frac{4-\sqrt x}{16x-x^2}=\frac{1}{128}$
Work Step by Step
$A=\lim\limits_{x\to16}\frac{4-\sqrt x}{16x-x^2}$
$A=\lim\limits_{x\to16}\frac{4-\sqrt x}{x(16-x)}$
Multiply both numerator and denominator by $4+\sqrt x$
We see that $(4-\sqrt x)(4+\sqrt x)=16-x$
since $(a-b)(a+b)=a^2-b^2$
Therefore,
$A=\lim\limits_{x\to16}\frac{(4-\sqrt x)(4+\sqrt x)}{x(16-x)(4+\sqrt x)}$
$A=\lim\limits_{x\to16}\frac{16-x}{x(16-x)(4+\sqrt x)}$
$A=\lim\limits_{x\to16}\frac{1}{x(4+\sqrt x)}$
$A=\frac{1}{16\times(4+\sqrt16)}$
$A=\frac{1}{128}$
