Finding limit of a quotient with two square roots: $\lim_{t\to 0}\frac{\sqrt{1+t}-\sqrt{1-t}}t$ Find
$$
\lim_{t\to 0}\frac{\sqrt{1+t}-\sqrt{1-t}}{t}. 
$$
I can't think of how to start this or what to do at all. Anything I try just doesn't change the function.

HINT $\ $ Use the same method in your prior question, i.e. rationalize the numerator by multiplying both the numerator and denominator by the numerator's conjugate $\rm\:\sqrt{1+t}+\sqrt{1-t}\:.$ Then the numerator becomes $\rm\:(1+t)-(1-t) = 2\:t,\:$ which  cancels with the denominator $\rm\:t\:,\:$ so $\rm\:\ldots$ 
More generally, using the same notation and method as in your prior question, if $\rm\:f_0 = g_0\:$ then
$$\rm \lim_{x\:\to\: 0}\ \dfrac{\sqrt{f(x)}-\sqrt{g(x)}}{x}\ = \ \lim_{x\:\to\: 0}\ \dfrac{f(x)-g(x)}{x\ (\sqrt{f(x)}+\sqrt{g(x)})}\ =\ \dfrac{f_1-g_1}{\sqrt{f_0}+\sqrt{g_0}}$$
In your case $\rm\: f_0 = 1 = g_0,\ \ f_1 = 1,\ g_1 = -1\:,\ $ so the limit $\: =\: (1- (-1))/(1+1)\:  =\: 1\:.$ 
Note again, as in your prior questions, rationalizing the numerator permits us to cancel the common factor at the heart of the indeterminacy - thus removing the apparent singularity.
