# Does limit exist at zero?

We know for a limit to exist the LHL and RHL should both be finite and equal or if they are infinite then both should tend towards positive infinity or both towards negetive infinity.

So my question is, if I define a function as $$F(x) = \begin{cases} |\ln(-x)|, & \text{for }x<0\\ 1/x,& \text{for }x>0 \end{cases}$$ Then does $$\lim_{x\to 0}F(x)$$ exist?

• Well, what happens when you test $F(x)$ and $F(-x)$ for small values of $x>0$? Say, $x=1/n$ where $n$ is an integer. Dec 19 '20 at 8:11
• @Michael it would give me the same result as earlier just in a mirrored way! I didn't get why you want me to do that. Dec 19 '20 at 8:59

$$\lim_{x\to 0} \lvert \ln (x) \rvert = \infty$$ $$\lim_{x\to 0^+} \frac{1}{x} = \infty$$
Whether this limit really exists is a matter of convention. Remember that $$\infty$$ is just a convenient shorthand for saying that the function grows indefinitely, i.e. it doesn't "tend" to any particular point which is what we define a limit to be. This is why most mathematicians would say that the limit of your function does not exist. In some fields, however, (and judging by the criteria you mention in your post this seems to be the case for you) infinite limits are said to exist. Your personal mileage may vary, so I suggest you stick with the definitions that have been provided to you and abide by that.
• @GouravM Suppose that you define $f(x) = \frac{-1}{|x|} : x \neq 0.$ Then, as $x \to 0,$ both ln$(x)$ and $f(x)$ go to $-\infty.$ Despite this, their limits are not equal at zero, because neither limit exists at $x=0$. $\infty$ is not a number but rather a symbol for unbounded growth. Dec 19 '20 at 8:25
• @GouravM $\varepsilon$ can't be infinite - it can be any positive number, no matter how large. Note that the definition of a (positively) infinite limit differs from that of a regular limit: we say $\lim_{x\to a} f(x) = \infty$ iff for all $\varepsilon >0$ there exists $N>0$ such that $0 < \lvert x - a\rvert < N \implies f(x)>M$. Dec 19 '20 at 10:08