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