Question on limits and infinity

Question

Just to clarify, the limit of $x \nearrow 0$ from the left of $1/x$, would be $-\infty$, and the limit of $x \searrow 0$ from the right of $1/x$, would be $+\infty$ right?

This is only true when its $1/x$ and not any other number over $x$? Sorry if this is confusing, are there certain formulas to know when the limit equals infinity?

Like the limit of $x \to 0$ of $1/x^2 = +\infty$

Thanks for any help, again sorry if this is confusing. I'm just trying to understand how to know when a limit equals infinity, rather then it does not exist.

Answer 1

To say that $\lim_{x \to a} f(a)$ exists, you have to check that both

$$\lim_{x \to a^+} f(a) \quad \textrm{ and } \quad \lim_{x \to a^+} f(a)$$

exist, and that they are equal.

To get a feel for these, it's perfectly fine to imagine plugging in a series of smaller and smaller values and seeing the result, i.e. a table of values.

For $\lim_{x \to 0} \frac{1}{x^2}$, let's start with $\lim_{x \to 0^+} \frac{1}{x^2}$. What do you get if you plug in $x = 0.1$, then $x = 0.01$, then $x = 0.001$? What does this indicate about the limit?

It suggests the limit is $+\infty$

Now, let's work with $\lim_{x \to 0^-} \frac{1}{x^2}$. Plug in $x = -0.1$, then $x = -0.01$, then $x = -0.001$. What limit do you get this time?

It suggests the limit is, once again, $+\infty$

Do these two limits agree? If so, then what can we say?

We can say that $\lim_{x \to 0} \frac{1}{x^2} = + \infty$

(note: we haven't rigorously proved that these limits are what we claim, but I'm guessing you're in a calculus course where this type of rigor isn't expected.)

Answer 2

The best way to see the intuitive difference is to examine what happens asymptotically as you approach zero. When $x \to 0^-$ this means that $x$ is negative, so the quantity $\frac1x$ is negative. Similarly, if $x \to 0^+$ then $x$ is positive and $\frac1x$ is positive. Using any other positive number in the numerator will give the same result, and a negative number in the numerator will reverse the signs. The formal way of checking that a limit diverges to positive infinity in the a one sided limit is as follows: Say we wanted to show that $$ \lim_{x\to c^+} f(x) = \infty,$$ we would want to show that $\forall N>0, \exists \delta>0$ such that if $x \in (c,c+\delta)$ then $f(x)>N$.

Now I think there is some ambiguity in whether or not a limit exists if it diverges. Typically, one says that a limit exists if both one sided limits exist and those limits are equal. In the case of $\frac1x$, both one sided limits diverge, but they diverge to "different infinities" while for $\frac1{x^2}$ both diverge to the "same infinity." I think some people exclude the possibility of infinite limits however.