Consider the function $F(x)$ whose graph is shown in Figure 1. If we take $x$ values closer and closer to 2.5, but less than 2.5, $F(x)$ gets closer and closer to 5. In other words, when $x$ approaches 2.5 through the values less than 2.5, $F(x)$ approaches 5. We express this by saying that “the limit of $F(x)$ as $x$ approaches 2.5 from the left is 5” or “the left-hand limit of $F(x)$ as $x$ approaches 2.5 is 5.” The notation for this is
\[\lim_{x\to2.5^{-}}F(x)=5\]
The minus sign that is written after 2.5 means $x$ approaches 2.5 from the left.
Figure 1: Graph of $y=F(x)$. |
Now consider the case in which $x$ takes on the values close to 2.5 but larger than 2.5. As $x$ approaches 2.5 from the right, $F(x)$ approaches 2. Symbolically we write
\[\lim_{x\to2.5^{+}}F(x)=2,\]
and say “the limit of $F(x)$ as $x$ approaches 2.5 from the right is 2” or “the right-hand limit of $F(x)$ as $x$ approaches 2.5 is 2.”
In this example, $F(x)$ is defined at $x=2.5$, but the value of $F(2.5)$ has no bearing on the left-hand or right-hand limit of $F(x)$. Even if we remove $x=2.5$ from the domain of $F(x)$ (that is, if $F(x)$ were not defined at $x=2.5$), the left-hand and right-hand limits would remain the same.
Similarly Do not forget the + sign. For example,\[\lim_{x\to2.5}f(x)\]means the limit of $f(x)$ as $x$ approaches 2.5 from both sides, but\[\lim_{x\to2.5^{+}}f(x)\] means the limit of $f(x)$ as $x$ approaches 2.5 from the right. Once again consider the sign function: By comparing the definitions of one-sided limits and regular (or two-sided) limits, we see the following is true. Theorem 1: ${\displaystyle \lim_{x\to a}f(x)}$ exists and is equal to $L$ if and only if ${\displaystyle \lim_{x\to a^{-}}f(x)}$ and ${\displaystyle \lim_{x\to a^{+}}f(x)}$ both exist and are equal to $L$. That is, \[\lim_{x\to a}f(x)=L\quad\Longleftrightarrow\quad\lim_{x\to a^{-}}f(x)=\lim_{x\to a^{+}}f(x)=L.\]
For instance, in the above example
Do not forget them and do not put them before $a$. For example,
\[\lim_{x\to-2.5}f(x)\] means the limit of $f(x)$ as $x$ approaches $-2.5$ from both sides. In general \[\lim_{x\to-2.5}f(x)\neq\lim_{x\to2.5^{-}}f(x).\]
$$ \lim_{x\to0^{-}}\text{sgn}(x)\neq \lim_{x\to0^{+}}\text{sgn}(x),$$
so ${\displaystyle \lim_{x\to0}\text{sgn}(x)}$ does not exist as we learned in the Section on the Concept of a Limit.