Diverging limits equating to infinity and negative infinity In trying to determine whether $\lim\limits_{x\rightarrow2}\frac{x}{x-2}=\infty$. I've found that the limit diverges and equates to both infinity and negative infinity. Does this mean that the limit does equal infinity or that it does not? What does a diverging limit mean?
 A: The convention is as follows, except for when it isn't.

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*$\lim_{x\to 2} f(x)=\infty$ means that for all $M$ there is some $\delta>0$ such that, for all $y\in (2-\delta,2+\delta)\setminus \{2\}$, $f(y)>M$. This is not the case for $f(x)=\frac x{x-2}$.


*$\lim_{x\to 2} f(x)=-\infty$ means that for all $M$ there is some $\delta>0$ such that, for all $y\in (2-\delta,2+\delta)\setminus \{2\}$, $f(y)<M$. This is not the case for $f(x)=\frac x{x-2}$.
It is true that $\lim_{x\to 2}\left\lvert \frac x{x-2}\right\rvert=\infty$, for the aforementioned reasons.
A: In order for $\ \displaystyle\lim_{x\to 2}\frac{x}{x-2}\ $ to exist [i.e. for the limit to converge], both: $\ \displaystyle\lim_{x\to 2^-}\frac{x}{x-2}\ $ and $\ \displaystyle\lim_{x\to 2^+}\frac{x}{x-2}\ $ must exist and be equal to the same finite real number. Else the limit does not converge, i.e. the limit diverges. If they both "equal $\ +\infty\ $", certainly it is true that the limit diverges (since $\ +\infty\ $ is not a finite real number), so we might say that the limit is "equal to $\ +\infty\ $", but this is usually considered an abuse of notation.
