Answer
$\lim\limits_{t\to0}(\frac{1}{t\sqrt{1+t}}-\frac{1}{t})=\frac{-1}{2}$
Work Step by Step
$A=\lim\limits_{t\to0}(\frac{1}{t\sqrt{1+t}}-\frac{1}{t})$
$A=\lim\limits_{t\to0}\frac{1-\sqrt{1+t}}{t\sqrt{1+t}}$
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 $t$)
$A=\frac{-1}{\sqrt{1+0}(1+\sqrt{1+0})}$
$A=\frac{-1}{2}$
