Delta Epsilon Question Help? I know that the delta epsilon can produce different correct answers depending on how you approach the question i have:
$\epsilon > 0$ be given
Show $|\sqrt{x^{2}+1}|$ is continuous at x=1.
So i Have:
$$ |\sqrt{x^{2}+1}-\sqrt{2}| \rightarrow \bigg|\sqrt{x^{2}+1}-\sqrt{2}*\frac{\sqrt{x^{2}+1}+\sqrt{2}}{\sqrt{x^{2}+1}+\sqrt{2}}\bigg| \rightarrow |x-1|\bigg|\frac{x+1}{\sqrt{x^{2}+1}+\sqrt{2}}\bigg|$$
Also as $|x-1|<\delta$ and so $x \in (0,2)$ 
Subbing in my values i get the min as $\frac{1}{1+\sqrt{2} \epsilon}$.
So therefore my value answer is $0 < \delta < min\bigg[1,\frac{1}{1+\sqrt{2} \epsilon }\bigg]$ would this be right as my minimum for delta as my lecturer in his notes times by 2 and has $2|x-1| < 2\delta < \epsilon$ so gets a min$\bigg[1,\frac{2}{\epsilon} \bigg]$.
So is my answer correct or do i need to work further to get to the answer in my lecture notes?
Any help would be appreciated thanks. 
 A: This is so fuzzy (and looks wrong), it is what you do in a draft. Here is how it is usually done (and the spirit behind it):
Let us consider $\epsilon > 0$.
Let us note $\delta = \min(\frac{1}{3}, \frac{\epsilon}{18\pi})$ (I decide to call this number $\delta$, because I like it, and I think it deserves a proper name)
Let us consider $x \in [1-\delta, 1+\delta]$ (I take any number in that set, I want to play a little with it)
Let us calculate $|\sqrt{x^2+1}-\sqrt{2}|$ (It's rainy outside and I don't have anything better to do, so let's do that. It is better than playing video games anyway)
$|\sqrt{x^2+1}-\sqrt{2}|\leq |\sqrt{x^2+1}-\sqrt{2}||\sqrt{x^2+1}+\sqrt{2}|$ (as $|\sqrt{x^2+1}+\sqrt{2}| \geq 1$ )
$ = x^2+1-2 \leq (1+\delta)^2-1 = 2\delta + \delta^2 \leq \frac{\epsilon}{18\pi}(2 + \delta) $ (since $\delta \leq \frac{\epsilon}{18\pi}$)
$\leq \frac{\epsilon}{6\pi} < \epsilon$ (since $\delta \leq 1$)
So $x \rightarrow \sqrt{x^2+1}$ is continuous in $1$ (Holy Saint Homer on a rainbow! This stuff I was calculating is smaller than $\epsilon $ all along! This means that my function is continuous! I really didn't expect that... )
Rq : You can take any $\delta$ up to $\min(1,\frac{\epsilon}{3})$ and do the same calculus.
