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.
 A: $$\frac{\sqrt{1+t}-\sqrt{1-t}}{t}$$
$$=\frac{\sqrt{1+t}-\sqrt{1-t}}{t}\cdot\frac{(\sqrt{1+t}+\sqrt{1-t})}{(\sqrt{1+t}+\sqrt{1-t})}$$
Note $(a-b)(a+b)=a^2-b^2$, so this is
$$\frac{(\sqrt{1+t})^2-(\sqrt{1-t})^2}{t(\sqrt{1+t}+\sqrt{1-t})}$$
$$=\frac{(1+t)-(1-t)}{t(\sqrt{1+t}+\sqrt{1-t})}$$
The top is $2t$, so this is
$$\frac{2}{\sqrt{1+t}+\sqrt{1-t}}.$$
To find this as $t\to0$ just plug in $t=0$, which gives
$$\frac{2}{\sqrt{1}+\sqrt{1}}=1.$$
A: This is probably not what you are expected to do (you probably are not supposed to know/use Taylor series at this point), but for the sake of information: this kind of limits are more easily solved with Taylor expansions. Knowing that around zero $\sqrt{1+x} = 1 + \frac{x}{2} + O(x^2)$, the result comes immediately.
A: 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.
