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.
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.