Calculating limits using limit laws I have been at this one for a while and I can't make it work. I am trying to evaluate the limit of $(\sqrt{1+t}-\sqrt{1-t})/t$ as t approaches 0. It seems pretty simple but I can't make it work. I have tried a lot of things but I think what I am meant to do is use the conjugate $(\sqrt{1+t}+\sqrt{1-t})$ which gives me $(1+t)-(1-t)/t(\sqrt{1+t}+\sqrt{1-t})$ 
Which then leaves me with $2t/t(\sqrt{1+t}+\sqrt{1-t})$
 A: You're on the right track with using the conjugate to get $$\frac{2t}{t(\sqrt{1+t}+\sqrt{1-t})}.$$  As has been said in the comments, this is equal to $$\frac{2}{(\sqrt{1+t}+\sqrt{1-t})}.$$  One way to think about that is $$\frac{2t}{t} = (2 \cdot t) \div t = 2 \cdot (t \div t) = 2 \cdot 1 = 2,$$ as in Austin Mohr's comment.  I tend to think about it slightly differently: $$\frac{2t}{t}=\left(2\cdot t\right)\cdot\frac{1}{t}=2\cdot\left(t\cdot\frac{1}{t}\right)=2\cdot1=2$$ (sticking to just multiplication, without division, using the Associative Property of Multiplication, and then using the fact that $t$ and $\frac{1}{t}$ are multiplicative inverses so that their product is $1$).  Or, for this particular problem, $$\begin{align}
\frac{2t}{t(\sqrt{1+t}+\sqrt{1-t})}
&=\left(2\cdot t\right)\cdot\left(\frac{1}{t}\cdot\frac{1}{\sqrt{1+t}+\sqrt{1-t}}\right)
\\
&=2\cdot\left(t\cdot\frac{1}{t}\right)\cdot\frac{1}{\sqrt{1+t}+\sqrt{1-t}}
\\
&=2\cdot 1\cdot\frac{1}{\sqrt{1+t}+\sqrt{1-t}}
\\
&=\frac{2}{\sqrt{1+t}+\sqrt{1-t}}.
\end{align}$$
