Evaluate the limit I am hung up on this limit: $\displaystyle\lim_{x\to0} \frac{\sqrt{1+x} + \sqrt{1-x}}{x}$
I must be missing something related to dealing with square roots but I can not for the life of me figure out what. 
Here is my work so far:
$\displaystyle\lim_{x\to0} \frac{\sqrt{1+x} + \sqrt{1-x}}{x} = \lim_{x\to0} \frac{(\sqrt{1+x} + \sqrt{1-x})(\sqrt{1+x} - \sqrt{1-x})}{x(\sqrt{1+x} + \sqrt{1-x})}$ 
$=\displaystyle \lim_{x\to0} \frac{\sqrt{1+x}^2 + (\sqrt{1+x} + \sqrt{1-x}) - (\sqrt{1+x} + \sqrt{1-x})-\sqrt{1-x}^2}{x(\sqrt{1+x} + \sqrt{1-x})}$ 
$= \displaystyle \lim_{x\to0} \frac{1+x - 1-x}{\sqrt{1+x} + \sqrt{1-x}}= 0$.
After this I end up with the answer 0, but I know that it should come out to 1. If someone could look over this and see where I am going wrong and point me in the right direction I will be eternally thankful!
 A: 
$$\lim\limits_{x\to0} \frac{\sqrt{1+x} + \sqrt{1-x}}{x}$$

Note that evaluating $\quad \displaystyle \lim \frac{\sqrt{1+x} + \sqrt{1-x}}{x}\;$ as $\;x\to 0\;$ gives you $\;2\;$ in the numerator, and $\;0\;$ in the denominator. So the task that ultimately remains is to evaluate the limits as $\;x \to 0^+\,$ and as $\;x \to 0^-$. 

Hint: (prior to post's edit): Regarding your algebraic manipulations, if you are attempting to "simplify" the expression to make the limit more evident: try multiplying numerator and denominator by $\;\sqrt{1+x} - \sqrt{1-x}$, and be careful with algebra!

$$\lim\limits_{x\to0} \frac{(\sqrt{1+x} + \sqrt{1-x})(\sqrt{1+x} - \sqrt{1-x})}{(\sqrt{1+x} - \sqrt{1-x})x}$$  $$= \lim_{x \to 0} \frac{2x}{(\sqrt{1+x} - \sqrt{1-x})x}$$
$$=\lim_{x \to 0} \frac{2}{\sqrt{1+x} - \sqrt{1-x}}$$
Now be sure to take the limit as $x \to 0^+$ and as $x\to 0^-$
A: I think that the question you meant to ask was to find $$\displaystyle\lim_{x\to0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x}.$$ This can be seen to be $1$ by binomial expansion of the terms in the numerator, and some cancellation. The limit of the expression you gave doesn't exist.
