Answer
$\lim\limits_{x\to-4}\frac{\sqrt{x^2+9}-5}{x+4}=\frac{-4}{5}$
Work Step by Step
$A=\lim\limits_{x\to-4}\frac{\sqrt{x^2+9}-5}{x+4}$
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)
$A=\frac{(-4)-4}{\sqrt{(-4)^2+9}+5}$
$A=\frac{-4}{5}$
