[tex]\displaystyle\lim_{x\to\infty}\frac{x^2+1}{x^2}=1[/tex]
means to say there is some number [tex]M[/tex] for which any choice of [tex]\varepsilon>0[/tex] such that
[tex]x>M\implies\left|\dfrac{x^2+1}{x^2}-1\right|<\varepsilon[/tex]
Working backwards, we have
[tex]\left|\dfrac{x^2+1}{x^2}-1\right|=\left|1+\dfrac1{x^2}-1\right|=\dfrac1{x^2}<\varepsilon[/tex]
[tex]\implies\sqrt{\dfrac1{x^2}}<\sqrt\varepsilon\implies\dfrac1{\sqrt\varepsilon}<|x|[/tex]
So, whenever [tex]x>M=\dfrac1{\sqrt\varepsilon}[/tex], we can always guarantee that [tex]\left|\dfrac{x^2+1}{x^2}-1\right|<\varepsilon[/tex].
