One-sided Limits¶

The input interval in the definition of limit for $\lim_{x \to a} f(x)$ is an open interval centered around $a$. These input intervals are always of the form $(a-\delta, a+\delta)$, where $\delta$ is a positive number.

For a right limit, we instead take the interval $(a, a+\delta)$. We denote right limits just like the usual (both-sided) limits, but with $x \to a^+$ instead of $x \to a$.

This is called the right limit, because on the number line, the input interval being used is the right half of the interval $(a-\delta, a+\delta)$.

Note that we don't need to say "any number in $I_x$ except $a$", like we do with both-sided limits; the number $a$ is not in any interval $(a, a+\delta)$. In the both-sided limit definition, we needed to exclude $a$ more explicitly, because it was in the center of an interval rather than being an end point.

We define the left limit similarly.

A consequence of these definitions is that if the function $f$ is defined like $$ f(x) = \begin{cases} \text{left part}, & \text{if $x < 4$}; \\ \text{center part}, & \text{if $x = 4$}; \\ \text{right part}, & \text{if $x > 4$}, \end{cases} $$ then $$ \begin{align} &\lim_{x \to 4^-} f(x) = \lim_{x \to 4^-} (\text{left part}), \\ &\lim_{x \to 4^+} f(x) = \lim_{x \to 4^+} (\text{right part}). \end{align} $$ The definition of the left limit doesn't use the values of $f(x)$ for $x \ge 4$ anywhere, so as far as it's concerned, $f(x)$ is just the left part. The right limit is similar. Also, notice that no limit with $x \to 4$, $x \to 4^-$ or $x \to 4^+$ uses the center part; limits are all about what happens around $4$, not at all about what happens at exactly $x=4$.

One-sided limits have all the nice properties that we have proved so far for both-sided limits; modifying the proofs to use one-sided limits instead of both-sided limits is easy.

Connection to both-sided limit¶

Suppose that $\lim_{x \to a} f(x) = L$. Now, for any output interval $I_y$ centered around $L$, we get an input interval $I_x=(a-\delta, a+\delta)$ centered around $a$ so that $$ f(\text{any number in $I_x$ except $a$}) \in I_y. $$ Because any number in $(a,a+\delta)$ is also in $I_x$, we get $$ f(\text{any number in $(a,a+\delta)$}) \in I_y, $$ which shows that $\lim_{x \to a^+} f(x) = L$. Similarly we get $\lim_{x \to a^-} f(x) = L$.

So, if the both-sided limit exists, then the one-sided limits exist too, and the values of all these limits are equal. However, if the one-sided limits exist, then the both-sided limit might still not exist! For example, let $J$ denote the "jumping function" given by $$ J(x) = \begin{cases} -1, & \text{if $x < 3$}; \\ +1, & \text{if $x \ge 3$}. \end{cases} $$ The both-sided limit $\lim_{x \to 3} J(x)$ doesn't exist, but the one-sided limits exist; they are $$ \lim_{x \to 3^-} J(x) = \lim_{x \to 3^-} -1 = -1 $$ and $$ \lim_{x \to 3^+} J(x) = \lim_{x \to 3^-} 1 = 1. $$ In this case, the one-sided limits are not equal. Let's see what happens to the both-sided limit if the one-sided limits are equal.

Suppose that $\lim_{x \to a^+} f(x) = L$ and $\lim_{x \to a^-} f(x) = L$. Let $I_y$ be any open interval centered around $L$. By the definition of the left and right limits, we get intervals $(a-\delta_1,a)$ and $(a,a+\delta_2)$ such that $$ \begin{align} &f(\text{any number in $(a-\delta_1,a)$}) \in I_y, \\ &f(\text{any number in $(a,a+\delta_2)$}) \in I_y. \end{align} $$

Let $s$ be the smaller of $\delta_1$ and $\delta_2$ (doesn't matter which if $\delta_1 = \delta_2$). Let $I_x=(a-s,a+s)$; this is an open interval centered around $a$.

The numbers on the "left half" of $I_x$ are in $(a-\delta_1,a)$ and the numbers on the "right half" are in $(a,a+\delta_2)$, so we get $$ f(\text{any number in $I_x$ except $a$}) \in I_y. $$

This result can be used to show more easily that the both-sided limit $\lim_{x \to 3} J(x)$ doesn't exist; if that limit existed, we would get $$ -1 = \lim_{x \to 3^-} J(x) = \lim_{x \to 3} J(x) = \lim_{x \to 3^+} J(x) = 1, $$ where the first and last equal signs come from the calculations above, and the rest of the equal signs come from the result. In general, the both-sided limit doesn't exist when the values of the left limit and the right limit differ.