# Proof limit value with $\epsilon/ \delta$

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?

Hint.

What you need is essentially the continuity of the function $$y\mapsto\sqrt{y}$$ at $$y=1$$.

So you need an estimate $$|\sqrt{y}-\sqrt{1}|<\epsilon$$

But observe that $$|\sqrt{y}-\sqrt{1}|=\frac{|y-1|}{\sqrt{y}+1}\le |y-1|$$

I believe this is an instance of the 'multiplying by 1' trick:

Let $$\epsilon > 0$$, $$\delta := \epsilon$$. Then, for $$x$$ such that $$|x|<\delta$$ we have:

$$|f(x) - 1| = |\sqrt{x+1} - 1| = |\frac{(\sqrt{x+1}-1)(\sqrt{x+1}+1)}{\sqrt{x+1}+1}| = |\frac{x}{\sqrt{x+1}+1}| \leq |x| < \delta = \epsilon$$