One-sided limits
Left-hand and right-hand limits
Sometimes a function approaches one value from the left and a different value from the right. Look at this function:
$f(x)=\begin{cases}x+1, & x<2\\3-x, & x\ge 2\end{cases}$
This means $f(x)=x+1$ when $x$ is less than $2$, and $f(x)=3-x$ when $x$ is $2$ or more.
As $x$ approaches $2$ from the left, $f(x)$ approaches $3$. As $x$ approaches $2$ from the right, $f(x)$ approaches $1$. We need a way to write each side separately. These are called one-sided limits.
The left-hand limit
We write
$\displaystyle\lim_{x\to a^-}f(x)=L$
and read it as "the limit of $f(x)$, as $x$ approaches $a$ from the left, is $L$." Here $x$ takes only values less than $a$.
The right-hand limit
We write
$\displaystyle\lim_{x\to a^+}f(x)=L$
and read it as "the limit of $f(x)$, as $x$ approaches $a$ from the right, is $L$." Here $x$ takes only values greater than $a$.
Note: The small signs in $a^-$ and $a^+$ only tell us the side. The minus sign means "from the left" (values less than $a$). The plus sign means "from the right" (values greater than $a$). They do not mean negative or positive numbers.
For the function above,
$\displaystyle\lim_{x\to 2^-}f(x)=3$ and $\displaystyle\lim_{x\to 2^+}f(x)=1$.
One-sided limits and the limit
The limit $\displaystyle\lim_{x\to a}f(x)$ uses both sides of $a$. So it is also called the two-sided limit. It is connected to the one-sided limits in a simple way:
$\displaystyle\lim_{x\to a}f(x)=L$ exactly when $\displaystyle\lim_{x\to a^-}f(x)=L$ and $\displaystyle\lim_{x\to a^+}f(x)=L$.
In other words:
- If the left-hand and right-hand limits are equal, the limit exists. It equals that common value.
- If the left-hand and right-hand limits are different, the limit does not exist.
For the function above, the one-sided limits at $x=2$ are $3$ and $1$. They are different, so $\displaystyle\lim_{x\to 2}f(x)$ does not exist.
Finding one-sided limits from a graph
Example 1: Use the graph of $g$ to find each value, if it exists.
(a) $\displaystyle\lim_{x\to -1^-}g(x)$ (b) $\displaystyle\lim_{x\to -1^+}g(x)$ (c) $\displaystyle\lim_{x\to -1}g(x)$ (d) $g(-1)$
(e) $\displaystyle\lim_{x\to 1^-}g(x)$ (f) $\displaystyle\lim_{x\to 1^+}g(x)$ (g) $\displaystyle\lim_{x\to 1}g(x)$ (h) $g(1)$
Solution:
At $x=-1$:
(a) From the left, the graph rises toward the open circle at height $2$. So $\displaystyle\lim_{x\to -1^-}g(x)=2$.
(b) From the right, the graph is the horizontal line at height $2$. So $\displaystyle\lim_{x\to -1^+}g(x)=2$.
(c) Both one-sided limits are $2$. So $\displaystyle\lim_{x\to -1}g(x)=2$.
(d) The solid point at $x=-1$ is at height $4$. So $g(-1)=4$. The limit and the function value are different.
At $x=1$:
(e) From the left, the graph is the horizontal line at height $2$. So $\displaystyle\lim_{x\to 1^-}g(x)=2$.
(f) From the right, the graph comes down toward the open circle at height $-2$. So $\displaystyle\lim_{x\to 1^+}g(x)=-2$.
(g) The one-sided limits are $2$ and $-2$. They are different, so $\displaystyle\lim_{x\to 1}g(x)$ does not exist.
(h) The solid point at $x=1$ is at height $2$. So $g(1)=2$.
Finding one-sided limits of a piecewise function
A piecewise function uses different formulas on different parts of its domain. To find a one-sided limit at the point where the formula changes:
- For the left-hand limit, use the formula for $x$ values less than $a$.
- For the right-hand limit, use the formula for $x$ values greater than $a$.
- In each case, substitute $x=a$ into that formula. (For simple formulas like these, substituting gives the limit. We will see why in computing limits.)
Example 2: Let $f(x)=\begin{cases}x^2, & x<1\\2x+1, & x\ge 1\end{cases}$
Find $\displaystyle\lim_{x\to 1^-}f(x)$, $\displaystyle\lim_{x\to 1^+}f(x)$, and $\displaystyle\lim_{x\to 1}f(x)$.
Solution:
For $x<1$, the formula is $x^2$. So,
$\displaystyle\lim_{x\to 1^-}f(x)=1^2=1$.
For $x>1$, the formula is $2x+1$. So,
$\displaystyle\lim_{x\to 1^+}f(x)=2(1)+1=3$.
The one-sided limits are $1$ and $3$. They are different, so $\displaystyle\lim_{x\to 1}f(x)$ does not exist.
Example 3: Let $h(x)=\begin{cases}x^2+1, & x<2\\3x-1, & x\ge 2\end{cases}$
Find $\displaystyle\lim_{x\to 2}h(x)$.
Solution:
Find each one-sided limit first.
$\displaystyle\lim_{x\to 2^-}h(x)=2^2+1=5$
$\displaystyle\lim_{x\to 2^+}h(x)=3(2)-1=5$
Both one-sided limits are $5$. So,
$\displaystyle\lim_{x\to 2}h(x)=5$.
Here the two formulas meet at the same point, so the graph has no jump at $x=2$.
More examples
Example 4: Find $\displaystyle\lim_{x\to 0^-}\dfrac{|x|}{x}$, $\displaystyle\lim_{x\to 0^+}\dfrac{|x|}{x}$, and $\displaystyle\lim_{x\to 0}\dfrac{|x|}{x}$.
Solution:
Recall the absolute value: $|x|=x$ when $x>0$, and $|x|=-x$ when $x<0$.
For $x<0$: $\dfrac{|x|}{x}=\dfrac{-x}{x}=-1$. So $\displaystyle\lim_{x\to 0^-}\dfrac{|x|}{x}=-1$.
For $x>0$: $\dfrac{|x|}{x}=\dfrac{x}{x}=1$. So $\displaystyle\lim_{x\to 0^+}\dfrac{|x|}{x}=1$.
The one-sided limits are $-1$ and $1$. They are different, so $\displaystyle\lim_{x\to 0}\dfrac{|x|}{x}$ does not exist.
Example 5: Find $\displaystyle\lim_{x\to 0^+}\sqrt{x}$.
Solution:
As $x$ approaches $0$ from the right, $\sqrt{x}$ approaches $\sqrt{0}=0$. So,
$\displaystyle\lim_{x\to 0^+}\sqrt{x}=0$.
The square root of a negative number is not a real number. So $\sqrt{x}$ is undefined for $x<0$, and there is no left-hand limit at $0$. At an endpoint of the domain like this, we use the one-sided limit from the side where the function is defined.