## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{h\to0^-}\frac{\sqrt{6}-\sqrt{5h^2+11h+6}}{h}=-\frac{11\sqrt6}{12}$$
$$\lim_{h\to0^-}\frac{\sqrt{6}-\sqrt{5h^2+11h+6}}{h}$$ To find one-side limit algebraically, we still apply the limit laws like for the two-side limits as normal. $$A=\lim_{h\to0^-}\frac{\sqrt{6}-\sqrt{5h^2+11h+6}}{h}$$ Here we need to multiply both nominator and denominator by $\sqrt{6}+\sqrt{5h^2+11h+6}$ to eliminate the $h$ in the denominator. - Nominator: $(\sqrt{6}-\sqrt{5h^2+11h+6})(\sqrt{6}+\sqrt{5h^2+11h+6})=6-(5h^2+11h+6)=6-5h^2-11h-6=-5h^2-11h$ - Denominator: $h(\sqrt{6}+\sqrt{5h^2+11h+6})$ Hence, $$A=\lim_{h\to0^-}\frac{-5h^2-11h}{h(\sqrt{6}+\sqrt{5h^2+11h+6})}$$ $$A=\lim_{h\to0^-}\frac{-5h-11}{\sqrt{6}+\sqrt{5h^2+11h+6}}$$ $$A=\frac{-5\times0-11}{\sqrt6+\sqrt{5\times0^2+11\times0+6}}$$ $$A=\frac{-11}{\sqrt6+\sqrt6}=\frac{-11}{2\sqrt6}=\frac{-11\sqrt6}{2\times6}=\frac{-11\sqrt6}{12}$$ Therefore, $$\lim_{h\to0^-}\frac{\sqrt{6}-\sqrt{5h^2+11h+6}}{h}=-\frac{11\sqrt6}{12}$$