Short and Sweet Calculus

## 2.2One-sided limits

Consider the function $$F(x)$$ whose graph is shown in Figure 2.2. 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.

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.

2.2. (left-hand limit) If we can make the values of $$f(x)$$ as close as we please to a number $$L$$ by taking $$x$$ sufficiently close to $$a$$ with $$\boldsymbol{x<a}$$, we say “the limit of $$f(x)$$ as $$x$$ approaches $$a$$ from the left is $$L$$” or “the left-hand limit of $$f(x)$$ as $$x$$ approaches $$a$$ is $$L$$” and write

$\lim_{x\to a^{-}}f(x)=L.$

Similarly

2.3. (right-hand limit) If we can make the values of $$f(x)$$ as close as we please to a number $$L$$ by taking $$x$$ sufficiently close to $$a$$ with $$\boldsymbol{x>a}$$, we say “the limit of $$f(x)$$ as $$x$$ approaches $$a$$ from the left is $$L$$” or “the left-hand limit of $$f(x)$$ as $$x$$ approaches $$a$$ is $$L$$” and write

$\lim_{x\to a^{-}}f(x)=L.$

Example 2.1. Consider the function $$y=\text{sgn}(x)$$ defined by ${\rm sgn}(x)=\left\{ \begin{tabular}{ll} \ensuremath{1} & if \ensuremath{x>0}\\ 0 & if \ensuremath{x=0}\\ \ensuremath{-1} & if \ensuremath{x<0} \end{tabular}\right.$ (a) Determine $$\lim_{x\to0^{-}}\text{sgn}(x)$$. (b) Determine $$\lim_{x\to0^{+}}\text{sgn}(x)$$.

Solution

The graph of $$y=\text{sgn}(x)$$ is shown below. (a) When $$x$$ is any negative number, the value of $$\text{sgn}(x)$$ is $$-1$$. Therefore $$\lim_{x\to0^{-}}\text{sgn}(x)}=-$$.
(b) When $$x$$ is positive, the value of $$\text{sgn}(x)$$ is $$1$$. Therefore, $$\lim_{x\to0^{+}}\text{sgn}(x)=1$$.

By comparing the definitions of one-sided limits and regular (or two-sided) limits, we see the following is true.

2.1. $$\lim_{x\to a}f(x)$$ exists and is equal to $$L$$ if and only if $$\lim_{x\to a^{-}}f(x)$$ and $$\lim_{x\to a^{+}}f(x)$$ 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$$\lim_{x\to0^{-}}\text{sgn}(x)}\neq\lim_{x\to0^{+}}\text{sgn}(x)$$, so $$\lim_{x\to0}\text{sgn}(x)$$ does not exist.