<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] You should proceed with your approach a little further. Note that in the last step the denominator is not less than $1/\sqrt{2}$ which itself exceeds $1/2$ and therefore $$\left|\dfrac{\dfrac{1}{1+x^2}-\dfrac{1}{2}}{\dfrac{1}{\sqrt{1+x^2}}+\dfrac{1}{\sqrt{2}}}\right|<2 \left|\frac{1}{1+x^2}-\frac{1}{2}\right|=\frac{|1-x^2|}{\sqrt{1+x^2}}$$ And the right most expression does not exceed $|1-x^2|$ and hence our job is done if we can ensure that $$|1-x^2|<\epsilon$$ or $$|x+1||x-1|<\epsilon $$ To control the factor $|x-1|$ you can note that $$|x+1|<1\implies |x-1|\leq |x+1|+|-2|<3$$ Thus if $0<|x+1|<1$ then we have $|x^2-1|<3|x+1|$ and the last expression can be made less than $\epsilon$ if $|x+1|<\epsilon/3$. It follows that we can choose $\delta=\min(1,\epsilon/3)$.
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