## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x\to-4}\frac{\sqrt{x^2+9}-5}{x+4}=\frac{-4}{5}$
$A=\lim\limits_{x\to-4}\frac{\sqrt{x^2+9}-5}{x+4}$ Multiply both the numerator and denominator by $\sqrt{x^2+9}+5$ - In the numerator, we find: $(\sqrt{x^2+9}-5)(\sqrt{x^2+9}+5)$ $=(x^2+9)-25$ (for $(a-b)(a+b)=a^2-b^2$) $=x^2-16$ $=(x-4)(x+4)$ (also for $(a-b)(a+b)=a^2-b^2$) Therefore, $A=\lim\limits_{x\to-4}\frac{(\sqrt{x^2+9}-5)(\sqrt{x^2+9}+5)}{(x+4)(\sqrt{x^2+9}+5)}$ $A=\lim\limits_{x\to-4}\frac{(x-4)(x+4)}{(x+4)(\sqrt{x^2+9}+5)}$ $A=\lim\limits_{x\to-4}\frac{x-4}{\sqrt{x^2+9}+5}$ (both $x+4$ get cancelled) $A=\frac{(-4)-4}{\sqrt{(-4)^2+9}+5}$ $A=\frac{-4}{5}$