$\lim_{t\to 0} \frac{(1+t)^{1/2} - (1-t)^{1/2}}{t}$ How can I find the limit of this? 
Should I use the conjugate pair? 
It doesn't seem logical that this has a limit, I started by trying to simplifying the function but I don't think you can simplify this any further.
$$
    \lim_{t\to 0}   \frac{(1+t)^{1/2} - (1-t)^{1/2}}{t}
$$
 A: $$\lim_{t\to 0}   \frac{\sqrt{1+t} - \sqrt{1-t}}{t}=\lim_{t\to 0}   \frac{\left(\sqrt{1+t} - \sqrt{1-t}\right)\left(\sqrt{1+t} + \sqrt{1-t}\right)}{\left(\sqrt{1+t} + \sqrt{1-t}\right)t}= \\
=\lim_{t\to 0}   \frac{{1+t} - {(1-t)}}{\left(\sqrt{1+t} + \sqrt{1-t}\right)t}=\lim_{t\to 0}   \frac{2t}{\left(\sqrt{1+t} + \sqrt{1-t}\right)t} = \\ 
=\lim_{t\to 0}   \frac{2}{\sqrt{1+t} + \sqrt{1-t}}=1$$
A: Hint (as mentioned in the comments). By multiplying by the conjugate you get
$$
\frac{[(1+t)^{1/2} - (1-t)^{1/2}]}{t}\cdot\frac{[(1+t)^{1/2} + (1-t)^{1/2}]}{[(1+t)^{1/2} + (1-t)^{1/2}]} = \frac{1+t - (1-t)}{t[(1+t)^{1/2} + (1-t)^{1/2}]}
$$
Now try to simplify this expression a bit and then look at $t\to 0$ again.
A: $$
    \lim_{t\to 0}   \frac{(1+t)^{1/2} - (1-t)^{1/2}}{t} 
= \lim_{t\to 0} \frac{((1+t)^{1/2} - (1 - t)^{1/2})((1+t)^{1/2}+(1-t)^{1/2})}{t((1+t)^{1/2} + (1-t)^{1/2})} 
$$
$$= \lim_{t\to 0} \frac{1 + t - (1 - t)}{t((1+t)^{1/2} + (1-t)^{1/2})} =\lim_{t\to 0} \frac{2t}{t((1+t)^{1/2} + (1-t)^{1/2})} $$
$$=\lim_{t\to 0}\frac{2}{((1+t)^{1/2} + (1-t)^{1/2})} = \frac{2}{2} = 1 $$
