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.
 A: 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 $\delta=\min(1,\epsilon/3)$.
A: $$| \frac{1}{\sqrt{x^2+1}} - \frac{1}{\sqrt 2}|$$ $$=\frac{|\sqrt{x^2+1}-\sqrt 2|}{\sqrt2 \sqrt{x^2+1}}.$$ Let $0<x-(-1)< \delta<1.$ We shall look at the numeraor and denominator of the expression above. First we deal with the denominator: $\sqrt{x^2+1}>1,$ so $$\frac{1}{\sqrt2 \sqrt{x^2+1}}<\frac{1}{\sqrt 2}$$ Now look at the numerator: $|x-(-1)|<\delta,$ so $-\delta<x-(-1)<\delta$ and $$-1-\delta<x<-1+\delta<0.$$ Hence $$\text{inequality 1: }(1-\delta)^2<x^2<(1+\delta)^2 $$ Now we take the right side of inequality 1: $$x^2<1+2\delta+\delta^2<1+3\delta$$
$$x^2+1<2+3\delta=2(1+\frac{3\delta}{2})$$ $$\sqrt{x^2+1}<\sqrt 2 \sqrt{1+\frac{3\delta}{2}}<\sqrt 2 (1+\delta) $$ because $(1+\delta)^2>1+2 \delta>1+\frac{3 \delta}{2}$ so $\sqrt{1+\frac{3 \delta}{2}}<1+\delta$ $$\text {Thus } \sqrt{x^2+1}<\sqrt 2+\sqrt 2   \delta$$ and hence 
$$\sqrt{x^2+1}-\sqrt 2<\sqrt 2   \delta$$
Now we take the left side of inequality 1: $$x^2>1-2\delta+\delta^2>1-2\delta$$ $$x^2+1>2-2\delta=2(1-\delta)$$
$$\sqrt{x^2+1}>\sqrt2 \sqrt{1-\delta}>\sqrt 2 (1-\delta)=\sqrt2 -\sqrt 2 \delta$$
$$\sqrt{x^2+1}-\sqrt2> -\sqrt 2 \delta$$ We see that $$|\sqrt{x^2+1}-\sqrt2|<\sqrt 2 \delta$$ Thus $$\frac{|\sqrt{x^2+1}-\sqrt 2|}{\sqrt2 \sqrt{x^2+1}}<\frac{\sqrt 2 \delta }{\sqrt 2}=\delta$$ Hence, if$ \delta=\min(1, \epsilon)$ then $$| \frac{1}{\sqrt{x^2+1}} - \frac{1}{\sqrt 2}|<\epsilon \text { whenever } |x-(-1)|<\delta.$$
A: 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.
A: Here is one way using the mean value theorem:
Let $\phi(x) = {1 \over \sqrt{x^2+1}}$ and note that $\phi'(x) = - {x \over \sqrt{(x^2+1)^3}}$ and $|\phi'(x)| \le |x|$.
You want to show $|\phi(x)-\phi(2)| < \epsilon$.
First pick $\delta = 1$. Then we know that $|x| \le 2+1 = 3$ and so the mean value theorem gives
$|\phi(x)-\phi(2)| < 3 |x-2|$ as long as $|x-2| < 1$.
But, we want it smaller. So pick $\delta' = {1 \over 3}  \epsilon$ (actually, $\delta' = \min(1,{1 \over 3}  \epsilon)$) then we have
$|\phi(x)-\phi(2)| < 3 \delta \le 3 {1 \over 3}  \epsilon= \epsilon$.
So, in this example, choose $\delta = {1 \over 3000}$.
