## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{t\to0}(\frac{1}{t\sqrt{1+t}}-\frac{1}{t})=\frac{-1}{2}$
$A=\lim\limits_{t\to0}(\frac{1}{t\sqrt{1+t}}-\frac{1}{t})$ $A=\lim\limits_{t\to0}\frac{1-\sqrt{1+t}}{t\sqrt{1+t}}$ Multiply both numerator and denominator by $1+\sqrt{1+t}$ We see that in the numerator: $(1-\sqrt{1+t})(1+\sqrt{1+t})=1-(1+t)=-t$ since $(a-b)(a+b)=a^2-b^2$ Therefore, $A=\lim\limits_{t\to0}\frac{(1-\sqrt{1+t})(1+\sqrt{1+t})}{t\sqrt{1+t}(1+\sqrt{1+t})}$ $A=\lim\limits_{t\to0}\frac{-t}{t\sqrt{1+t}(1+\sqrt{1+t})}$ $A=\lim\limits_{t\to0}\frac{-1}{\sqrt{1+t}(1+\sqrt{1+t})}$ (divide numerator and denominator by $t$) $A=\frac{-1}{\sqrt{1+0}(1+\sqrt{1+0})}$ $A=\frac{-1}{2}$