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

• You probably mean $0 < e \leq 1$ ? – Winther Nov 28 '20 at 18:04

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 $$x, then $$|\frac{1}{1+x^2}-0|<\epsilon$$, that is $$x 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 $$K=-\sqrt{\frac{1}{e}-1}$$.
• 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 $e \gt 1$ – Rajkumar Kumawat Nov 29 '20 at 16:35
• 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 $x<K$. – boaz Nov 29 '20 at 20:34