I have to find the limit value for $f(x)=\sqrt{1+x}$ for $x \rightarrow0$. And then show with $\epsilon /\delta$ that I have found the right limit value. I have found the limit value to $\sqrt{1+x} \rightarrow1$ for $x \rightarrow0$, by: $$\lim_{x \rightarrow 0} \sqrt{1+x} =\sqrt{\lim_{x \rightarrow 0} (1+x)}=\sqrt{\lim_{x \rightarrow 0} (1)+\lim_{x \rightarrow 0}(x)}=\sqrt{1+\lim_{x \rightarrow 0}(x)}=1$$
But I got a problem when I have to show it by $\epsilon /\delta$. When I write: $$|f(x)-1|\Rightarrow|\sqrt{1+x}-1|$$ I can't see what I can use as a bound easier to find a $\delta$. Can anyone help me?