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$x^2> \frac1e-1 $ and $e \gt 0$ $\implies x\lt -\sqrt{\lvert \frac 1e-1\rvert} $ or $ x\gt \sqrt{\lvert \frac 1e-1\rvert} $

In my text book I have come across this implication. I am not able to justify its validity. In my opinion It should be as below

Given : $x^2> \frac1e-1 $ and $e \gt 0$

Case 1: $(\frac 1e-1) \gt0$ then $x^2> \frac1e-1 \implies \lvert x\rvert^{2} \gt$ $(\sqrt{\lvert \frac 1e-1\rvert})^2 \implies \lvert x\rvert \gt (\sqrt{\lvert \frac 1e-1\rvert})$

Case 2: $(\frac 1e-1) \lt0$ then $x^2> \frac1e-1 \implies -x^2\lt -(\frac1e-1) \implies -x^2\lt \lvert \frac 1e-1\rvert$

I am not able to reach the given implication.

Any help towards this will be appreciated.

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  • $\begingroup$ You probably mean $0 < e \leq 1$ ? $\endgroup$ – Winther Nov 28 '20 at 18:04
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I think that this statement is false: Take for example $e=4/3$, so the statement is $$ x^2>-\frac{1}{4}\ \Rightarrow\ x<-\frac{1}{2}\ \text{or}\ x>\frac{1}{2} $$
$x=0$, for instance, is a counterexample.

Regarding the reference: The proof is not quite accurate. Given $\epsilon>0$ you need to prove that there is $K$ such that if $x<K$, then $|\frac{1}{1+x^2}-0|<\epsilon$, that is $$ x<K\Rightarrow \frac{1}{1+x^2}<\epsilon. $$ It should be separated into two cases: if $1<\epsilon$, then you may take $K=0$, since $\frac{1}{1+x^2}\leq 1<\epsilon$ for every $x$. If $0<\epsilon\leq 1$, then as indicate in the reference, you may take $K=-\sqrt{\frac{1}{e}-1}$.

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  • $\begingroup$ Yes your counterexample is enough to disprove it. I have added a reference to the statement. Can you please have a look at it and guide me if I am missing some point here. $\endgroup$ – Rajkumar Kumawat Nov 29 '20 at 4:31
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    $\begingroup$ See my additional comments @Rajkumar Kumawat. $\endgroup$ – boaz Nov 29 '20 at 7:00
  • $\begingroup$ Thanks for your detailed comment. But according to the definition as given in the reference above K is required to be positive. So we can't have $K=0$ right? Also it looks like we can have $K=e$ for the case $ e \gt 1$ $\endgroup$ – Rajkumar Kumawat Nov 29 '20 at 16:35
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    $\begingroup$ Actually, it does not matter. You may equivalently, say that $\lim_{x\to-\infty}f(x)=L$ iff for every $\epsilon$ , there exist a $K$ (positive or negative) such that $x<K$ implies $|f(x)-L|<\epsilon$. If the choice $K=0$ is still bother you, take any other $K$. It will be still true since $\frac{1}{1+x^2}<\epsilon$, as I indicated, is true for all $x$, regardless the condition $x<K$. $\endgroup$ – boaz Nov 29 '20 at 20:34

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