## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x\to16}\frac{4-\sqrt x}{16x-x^2}=\frac{1}{128}$
$A=\lim\limits_{x\to16}\frac{4-\sqrt x}{16x-x^2}$ $A=\lim\limits_{x\to16}\frac{4-\sqrt x}{x(16-x)}$ 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, $A=\lim\limits_{x\to16}\frac{(4-\sqrt x)(4+\sqrt x)}{x(16-x)(4+\sqrt x)}$ $A=\lim\limits_{x\to16}\frac{16-x}{x(16-x)(4+\sqrt x)}$ $A=\lim\limits_{x\to16}\frac{1}{x(4+\sqrt x)}$ $A=\frac{1}{16\times(4+\sqrt16)}$ $A=\frac{1}{128}$