## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to3}\frac{\sqrt{x+6}-x}{x^3-3x^2}=-\frac{5}{54}$$
$$A=\lim\limits_{x\to3}\frac{\sqrt{x+6}-x}{x^3-3x^2}$$ Multiply both numerator and denominator by $\sqrt{x+6}+x$, we have: - Numerator: $(\sqrt{x+6}-x)(\sqrt{x+6}+x)=(x+6)-x^2=-(x^2-x-6)=-(x-3)(x+2)$ - Denominator: $(x^3-3x^2)(\sqrt{x+6}+x)=x^2(x-3)(\sqrt{x+6}+x)$ Therefore, $$A=\lim\limits_{x\to3}\frac{-(x-3)(x+2)}{x^2(x-3)(\sqrt{x+6}+x)}$$$$A=-\lim\limits_{x\to3}\frac{x+2}{x^2(\sqrt{x+6}+x)}$$$$A=-\frac{3+2}{3^2\times(\sqrt{3+6}+3)}$$$$A=-\frac{5}{54}$$