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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.

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Well, $\frac{17}{x}$ would give exactly the same conclusion, and $\frac{-3}{x}$ would give a similar conclusion, with signs reversed. –  André Nicolas Sep 10 '13 at 21:46
    
Well, what happened with $\frac{1}{x}$ was that it approached one thing from one direction, but it approached something else when coming from the other direction. Now, what might you expect if the "thing you approach" is the same on both sides? –  Dennis Meng Sep 10 '13 at 21:48

2 Answers 2

up vote 1 down vote accepted

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.)

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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.

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