# Is this formulation of the statement that a function diverges to $-\infty$ in terms of sequences consistent?

I am asked to answer a question in which I am told that $$f:\mathbb{R}\to\mathbb{R}$$ is such that $$$$\lim_{x\to+\infty}f=\lim_{x\to-\infty}=-\infty$$$$ And, in order to make a start in answering the given question, I am trying to formulate a statement which is equivalent to the given statement, but in terms of sequences of real numbers. I know that, for any real-valued function $$f$$ defined on $$(a,b)$$, except possibly $$c\in(a,b)$$, we have that $$\lim_{x\to c}f(x)=l\iff\forall\langle x_n\rangle~\mathrm{such~that}~x_n\in(a,b),~x_n\neq c,~\forall n\in\mathbb{N},~x_n\to c\implies f(x_n)\to l$$ For this reason, it feels intuitive to claim that $$\lim_{x\to +\infty}f(x)=-\infty\iff\forall\langle x_n\rangle~\mathrm{such~that}~x_n\in(a,+\infty),~\forall n\in\mathbb{N},~x_n\to +\infty\implies f(x_n)\to -\infty$$ and $$\lim_{x\to -\infty}f(x)=-\infty\iff\forall\langle x_n\rangle~\mathrm{such~that}~x_n\in(-\infty,b),~\forall n\in\mathbb{N},~x_n\to -\infty\implies f(x_n)\to -\infty$$ Although I'm not sure whether or not this is rigorous. Any help is appreciated.

• The above is correct - however, you can dispense with $a,b$ and can use that for any sequence $x_n \to \pm \infty$ it follows that $f(x_n) \to -\infty$. If needed you can even juxtapose the above and use that for any unbounded sequence $|x_n| \to \infty$ (so could have subsequences going to both plus and minus infinities) the same conclusion holds – Conrad Dec 3 at 0:26