existence of limit when both RHL and LHL approachees Positive infinity We know that if $$\lim_{x \to a^+}f(x)=\lim_{x \to a^-}f(x)=L$$ Then $$\lim_{x \to a}f(x) =L$$ if $L$ is finite
But if $$\lim_{x \to a^+}f(x) \to +\infty$$ and
$$\lim_{x \to a^-}f(x) \to +\infty$$
Can we say $$\lim_{x \to a} f(x)$$ Does not exists since we cannot compare two infinities.
 A: Having $$\lim_{x \to a^+} f(x)=\infty$$ alone is enough to conclude that the limit does not exist, since otherwise there is a finite $L$ such that $$\lim_{x \to a} f(x)=L,$$ which implies that $$\lim_{x \to a^+} f(x)=L.$$
A: In general, we only say a limit 'exists' when it is finite. When we say $\lim_{x\to a}f(x)=+\infty$ we mean that
$$(\forall M>0)\,(\exists \delta > 0)\,(\forall x \,\text{ with }\, 0<|x-a|<\delta)\,\, f(x)>M \tag{1}$$
In other words, $f$ becomes arbitrarily large near $a$ — all of $f$. Contrast this with '$f$ attains arbitrarily large value near $a$'.

When we say $\lim_{x\to a^+}f(x)=+\infty$ we mean that
$$(\forall M>0)\,(\exists \delta > 0)\,(\forall x \,\text{ with }\, 0<|x-a|<\delta\,\text{ and }\, x>a)\,\, f(x)>M\tag{2}$$
In other words, $f$ becomes arbitrarily large in a right-neighborhood of $a$ — all of $f$, once again.

When we say $\lim_{x\to a^-}f(x)=+\infty$ we mean that
$$(\forall M>0)\,(\exists \delta > 0)\,(\forall x \,\text{ with }\, 0<|x-a|<\delta\,\text{ and }\, x<a)\,\, f(x)>M\tag{3}$$
In other words, $f$ becomes arbitrarily large in a left-neighborhood of $a$ — all of $f$, once again.

Can you see how statements $(2)$ and $(3)$, combined, imply statement $(1)$?
A: 
Can we say $\lim \limits_{x\rightarrow a}f(x)$ does not exists since we cannot compare two infinities?

Yes, we can say that it does not exist at a point $a$ because plus and minus infinities are different behaviors (function growing without bound/decreasing without bound). Moreover, limits of functions that end up in plus or minus infinity actually do not exist either. However, we do write that a limit is equal to $\pm\infty$, but that's only done for the purposes of describing the way a function behaves. It's just supposed to be more informative that way, meaning it gives us more information about a function's behavior near a particular point.
A: You can say that if you only want to deal with finite limits. But there is a sense in which such functions, for example, $$\frac{1}{x^2}$$ at $x=0$ behave regularly (though grow beyond any bound) at such points. In this case we may then say that the function is infinite at $a,$ so e.g., we may say that $1/x^2$ is infinite at $0.$
