One-sided Limits¶

In our definition of limit, "$x$ close enough to $a$" means that $x$ is in some interval centered around $a$, but $x \ne a$. This means that the interval is $(a-r, a+r)$, where $r$ is a positive number.

For a right limit, we instead take the interval $(a, a+r)$, In other words, we only consider the $x$ values where $x > a$. 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 interval being used is the right half of the original $(a-r, a+r)$.

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. For example, replace "close enough to $a$" with "close enough to $a$ on the right side".

Connection to both-sided limit¶

Suppose that $\lim_{x \to a} f(x) = L$. This means that for any tolerance $t > 0$, we have $f(x) \approx L$ with tolerance $t$ when $x$ is close enough to $a$. Because this works with any $x$ that is close to $a$, it obviously works with any $x$ that is close to $a$ on the right side. Hence we also have $$ \lim_{x \to a^+} f(x) = L, $$ and similarly $$ \lim_{x \to a^-} f(x) = L. $$ 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 $$ J(x) = \begin{cases} -1, & \text{if $x < 3$}; \\ +1, & \text{if $x \ge 3$}. \end{cases} $$

This function "jumps" at $x=3$:

The both-sided limit $\lim_{x \to 3} J(x)$ doesn't exist, because the function would need to be close to its limit on both sides of $3$. However, 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 $\lim_{x \to a^-} f(x) = L$. Then for any tolerance $t > 0$, we have $f(x) \approx L$ with tolerance $t$ if $x$ is close enough to $a$ on the left side, or on the right side. The corresponding intervals look like this:

Let $s$ be the smaller of $r_1$ and $r_2$ (doesn't matter which if $r_1 = r_2$).

Using the interval $(a-s,a+s)$ for the concept of close enough, we see that $f(x) \approx L$ with tolerance $t$ when $x$ is close enough to $a$, regardless of which side it's on. So if the left limit and right limit are both $L$, the usual both-sided limit is also $L$.

This result is often used to show that a both-sided limit doesn't exist. For example, $\lim_{x \to 3} J(x)$ cannot exist, because the left limit is $-1$ and the right limit is $1$. If the both-sided limit existed, it would need to be equal to $-1$ and equal to $1$, which is not possible.