Proof limit value with $\epsilon/ \delta$

Question

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?

Answer 1

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

Answer 2

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:

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