<s>[INST] Find delta for the limit I'm having difficulty to solve this problem: 
I know that ${\displaystyle \lim_{x\to a} f(x) = L}$ means for every $\varepsilon > 0$, there exists $\delta > 0$ such that $|f(x) - L| < \varepsilon$ whenever $0 <|x-a| < \delta$. 
I need to find $\delta$ when $\varepsilon = 0.001$ for ${\displaystyle \lim_{x \to -1} \frac{1}{\sqrt{x^2+1}} = \frac{1}{\sqrt 2}.}$ 
I've started as this:
$$ \left
|\frac{1}{\sqrt{x^2+1}} - \frac{1}{\sqrt 2}\right| 
= \left|\left(\frac{1}{\sqrt{x^2+1}} - \frac{1}{\sqrt 2}\right)\frac{\left(\frac{1}{\sqrt{x^2+1}} + \frac{1}{\sqrt 2}\right)}{\left(\frac{1}{\sqrt{x^2+1}} + \frac{1}{\sqrt 2}\right)} \right| = \left| \frac{\frac{1}{x^2+1} - \frac 12}{\left(\frac{1}{\sqrt{x^2+1}} + \frac{1}{\sqrt 2}\right)}
\right| < \epsilon.$$
But I do not know how can I finish solving this to find x.
 [/INST] Firstable you can make $|1+x| < 1 \implies |x| -1 =|x| - |1| \le |1+x| < 1$. So:
$|x| < 2$. Now call the expression just before the $\epsilon$ in your work above $D$, then $D \le \dfrac{|1-x^2|}{2\cdot \frac{1}{\sqrt{2}}} \le |1-x^2|= |1+x||1-x|\le |1+x|(1+|x|) < 3|1+x|< \epsilon $ if $1+x| < \frac{\epsilon}{3}$. Thus choose $\delta = \min(1,\frac{\epsilon}{3})$,and you're done.
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