$\displaystyle\lim _{x\to 0}\frac{\sqrt{1+x+x²}-1}{\sqrt{1+x}-\sqrt{1-x}}$ First I need to prove that this limit; $\displaystyle\lim _{x\to 0}\frac{\sqrt{1+x+x^2}-1}{\sqrt{1+x}-\sqrt{1-x}}$ converges, then I have to find its limit. Now I don't know how to prove that it converges (these epsilon proofs are still something that I'm trying to learn). And to actually find the limit; I tried to rewrite it by multiplying by $\displaystyle{\sqrt{1+x}+\sqrt{1-x}\over \sqrt{1+x}+\sqrt{1-x}}$ but I didn't get any further..
Edit: I know that it converges to $1/2$, but that's what wolfram alpha says :|
 A: The standard way is to transform the limit into
$$
\lim _{x\to 0}
  \frac{(\sqrt{1+x+x^2}-1)(\sqrt{1+x+x^2}+1)}
       {(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}
  \cdot
  \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x+x^2}+1}
$$
that becomes
$$
\lim_{x\to0}
  \frac{x+x^2}{2x}
  \cdot
  \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x+x^2}+1}
=
\lim_{x\to0}
  \frac{1+x}{2}
  \cdot
  \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x+x^2}+1}
$$
The alternative is to use Taylor expansions:
\begin{gather}
\sqrt{1+x+x^2}=1+\frac{1}{2}(x+x^2)+o(x)=1+\frac{1}{2}x+o(x)
\\
\sqrt{1+x}=1+\frac{1}{2}x+o(x)
\\
\sqrt{1-x}=1-\frac{1}{2}x+o(x)
\end{gather}
So the limit is
$$
\lim_{x\to0}\frac{1+\frac{1}{2}x+o(x)-1}{(1+\frac{1}{2}x+o(x))-(1-\frac{1}{2}x+o(x))}=\frac{\frac{1}{2}x+o(x)}{x+o(x)}=\frac{1}{2}
$$
A: $$\frac{\sqrt{1+x+x^2}-1}{\sqrt{1+x}-\sqrt{1-x}}=\frac{x+x^2}{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x+x^2}+1)}=$$   $$=\frac{x+x^2}{\sqrt{1+x}-\sqrt{1-x}}*\frac{1}{\sqrt{1+x+x^2}+1}$$ It follows
$$\lim _{x\to 0}\frac{\sqrt{1+x+x^2}-1}{\sqrt{1+x}-\sqrt{1-x}}=\lim _{x\to 0}\frac{x+x^2}{\sqrt{1+x}-\sqrt{1-x}}*\lim _{x\to 0}\frac{1}{\sqrt{1+x+x^2}+1}$$ The second factor is clearly equal to $\color{red}{\frac 12}$ and $$\lim _{x\to 0}\frac{x+x^2}{\sqrt{1+x}-\sqrt{1-x}}=\lim _{x\to 0}\frac{(x+x^2)(\sqrt{1+x}+\sqrt{1-x})}{2x}=\lim _{x\to 0}\frac{(1+x)(\sqrt{1+x}+\sqrt{1-x})}{2}=\frac 22=1$$ Thus the asked limit is equal to $$1\cdot \frac 12=\color{red}{\frac 12}$$
