E-d with infinity

Question

I saw that: \begin{align*} N &>0\\ x &\to \infty \implies x > N \\ x &\to -\infty \implies x<N\\ \end{align*} But \begin{align*} M &>0\\ f(x) &\to \infty \implies f(x) > M \\ f(x) &\to -\infty \implies f(x)<-M\\ \end{align*} why isn't it $$f(x) \to -\infty \implies f(x) < M$$ or $$ x\to -\infty \implies x<-N$$ Do both work. If not/yes then why?

Answer 1

$x \to \infty \implies x \gt N $( here , $N\gt 0 $ is taken as sufficiently large.) That means, as , $x \to \infty $ , so, however large, $N\gt 0 $ , we take , we must always have such $x\gt 0 $ that $x\gt N $

Whenever, $x \to -\infty \implies x \lt -N $( here , $N\gt 0 $ is taken as sufficiently large.) That means, as , $x \to -\infty $ , so, however large, $N\gt 0 $ , we take , we must always have such $x\lt 0 $ that $x\lt -N $

Similar case also for $f(x)$ .