Answer
$\lim_{x\to\infty}f(x)=L$ means that for every number $\epsilon\gt0$, there exists a corresponding number $M$ such that for all $x$,
$$x\gt M\Rightarrow |f(x)-L|\lt\epsilon$$
$\lim_{x\to-\infty}f(x)=L$ means that for every number $\epsilon\gt0$, there exists a corresponding number $N$ such that for all $x$,
$$x\lt N\Rightarrow |f(x)-L|\lt\epsilon$$
Work Step by Step
1) $\lim_{x\to\infty}f(x)=L$ means that for every number $\epsilon\gt0$, there exists a corresponding number $M$ such that for all $x$,
$$x\gt M\Rightarrow |f(x)-L|\lt\epsilon$$
$\lim_{x\to-\infty}f(x)=L$ means that for every number $\epsilon\gt0$, there exists a corresponding number $N$ such that for all $x$,
$$x\lt N\Rightarrow |f(x)-L|\lt\epsilon$$
2) Examples:
We examine the function $y=\frac{1}{x}$
As $x\to\pm\infty$, $f(x)$ approaches $0$. So $\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=0$
$\lim_{x\to\infty}f(x)=0$ means that for every number $\epsilon\gt0$, there exists a corresponding number $M$ such that for all $x$,
$$x\gt M\Rightarrow |f(x)|\lt\epsilon$$ $$x\gt M\Rightarrow |\frac{1}{x}|\lt\epsilon$$ $$x\gt M\Rightarrow -\epsilon\lt\frac{1}{x}\lt\epsilon$$ $$x\gt M\Rightarrow x\in(-1/\epsilon,0)\cup(1/\epsilon,\infty)$$
So if we take $M=\frac{1}{\epsilon}$, since $x\gt M$, then it is always the case that $|f(x)|\lt\epsilon$.
For example, if $\epsilon=2$, then for all $x\gt\frac{1}{2}$, we have $|\frac{1}{x}|\lt2$.
Similary, $\lim_{x\to-\infty}f(x)=0$ means that for every number $\epsilon\gt0$, there exists a corresponding number $N$ such that for all $x$,
$$x\lt N\Rightarrow |f(x)|\lt\epsilon$$ $$x\lt N\Rightarrow x\in(-1/\epsilon,0)\cup(1/\epsilon,\infty)$$
So if we take $N=\frac{-1}{\epsilon}$, since $x\lt N$, then it is always the case that $|f(x)|\lt\epsilon$.
For example, if $\epsilon=2$, then for all $x\lt\frac{-1}{2}$, we have $|\frac{1}{x}|\lt2$.
