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# One sided limits – Definition, Techniques, and Examples

What happens when the graph of a function contains discontinuities such as piecewise and rational functions? The function’s behavior may differ as the function approaches from the left or the right, so we instead inspect its one-sided limits.

*One-sided limits are the function’s limits as they approach a restricted value or side of the function. *

This article will review discontinuities and how they affect the graph’s limit as it approaches from the left or right of $x = a$. We’ll also introduce you to the techniques of finding the limits of a function from the left and from the right.

## What is a one sided limit?

A one-sided limit can either be of the form $\lim_{x \rightarrow a^{-}} f(x)$ or $\lim_{x \rightarrow a^{+}} f(x)$. The plus or minus sign following $a$ indicate the region or interval that we want to observe.

From the notations alone, we can see that one-sided limits are the limits of the function as it approaches a certain value and interval.

One-sided limits are helpful when we want to check for discontinuities and if we want to confirm if a limit exists. Let’s observe the piecewise function, $f(x)=\left\{\begin{matrix}x,&x < 0\\ 5 – x,&x \geq 0\end{matrix}\right.$, and the behavior of $x$ as it approaches $0$.

This piecewise function is a good example of why we study one-sided limits. As $x$ approaches $0$ from the left, we can see that the limit of $f(x)$ is equal to $0$, and when $x$ approaches $0$ from the right, the limit becomes $5$.

This shows that it is possible for the limit from each side of the function may be different.

### What does the limit from the left mean?

Let’s dive into understanding what $\lim_{x \rightarrow a^{-}} f(x)$ represent. When we see $\lim_{x \rightarrow a^{-}}$ , this means that the limit is evaluated from the left of $a$.

When we say limit from the left of $a$, we want to find the limit of $f(x)$ as $x<a$ without letting $x$ be $a$.

For the piecewise function, $f(x)=\left\{\begin{matrix}x,&x < 0\\ 5 – x,&x \geq 0\end{matrix}\right.$, $\lim_{x \rightarrow 0^{-} }f(x)$ represents the limit of the function as $x$ approaches zero coming from the values that are less than 0.

### What does the limit from the right mean?

Similarly, whenever we see $\lim_{x\rightarrow a^{+} }$, this means that the limit is evaluated from the right of $a$ or when $x<a$ without letting $x$ be $a$.

Using the same example, $f(x)=\left\{\begin{matrix}x,&x < 0\\ 5 – x,&x \geq 0\end{matrix}\right.$, $\lim_{x \rightarrow 0^{+}} f(x)$ represents the limit of $f(x) as $x$ approaches zero coming from the values that are greater than 0.

## How to find one sided limits?

Since the only difference between one-sided and regular limits are the intervals we’re working with, make sure to review your knowledge on evaluating limits as we might need some of the techniques to find one-sided limits as well.

### Finding one-sided limits from a graph

We can find either $\lim_{x \rightarrow a^{-}} f(x)$ or $\lim_{x \rightarrow a^{+}} f(x)$ by inspecting the graph’s behavior when $x < a$ or when $x > a$, respectively.

The graph of the piecewise function, $f(x)=\left\{\begin{matrix}3x,&x < 0\\ 12 – 2x,&x \geq 0\end{matrix}\right.$, is as shown above.

If we want to find $\lim_{x \rightarrow 0^{-}} f(x)$, we’ll look at the value that the interval $x<0$ approaches. Looking at the graph, we can see that from the left of $0$, the value approaches $0$.

Similarly, to find $\lim_{x \rightarrow 0^{+}} f(x)$, let’s inspect the value of $f(x)$ that the values to the right of $0$ approach. The graph shows that the limit from the right would be equal to $12$ instead.

### Finding one-sided limits algebraically

When given the expression for the function, and you need to find the limit as $x$ approaches $a$ from the left or $a$ from the right, it helps to keep in mind which values are being satisfied given a certain interval.

- Check which of the expressions will the values of $x<a$ belong to.
- Evaluate the limits at $x = a$ and for the corresponding expression of the function.

Let’s try finding $\lim_{x \rightarrow 2^{-}} f(x)$ and $\lim_{x \rightarrow 2^{+}} f(x)$ given the piece function, $f(x)=\left\{\begin{matrix}-x^2,& 0 \leq x\leq 2\\ 2x+ 1,&2 <x\leq 4\end{matrix}\right.$.

If we want to find the limit from the left or $\lim_{x \rightarrow 2^{-}} f(x)$, first check which of the two expressions would we be using for values that are from the left of $2$?

$f(x)=\left\{\begin{matrix}\color{blue}-x^2,& \color{blue}0 \leq x \leq 2\\ 2x + 1,&2 < x \leq 4\end{matrix}\right.$

We can see that we have $f(x) = -x^2$ when we want the values from the left of $2$, so $\lim_{x \rightarrow 2^{-}} f(x) = \lim_{x \rightarrow 2^{-}} –x^2$. Since $-x^2$ is a polynomial, we simply substitute $x=-2$ into $-x^2$.

$\begin{aligned}\lim_{x \rightarrow 2^{-}} f(x)&= f(2), \text{where } f(x) = -x^2\\ &= -(2)^2\\&=-4 \end{aligned}$

We can apply the same thought process for finding the value of $\lim_{x \rightarrow 2^{-}} f(x)$.

$\begin{aligned}\lim_{x \rightarrow 2^{+}} f(x)&= f(2), \text{where } f(x) = 2x + 1\\ &= 2(2) + 1\\&= 5 \end{aligned}$

This means that the limit of $f(x)$ from the left is $-4$ while the limit of $f(x)$ from the right is $5$.

## How to find confirm if a function has a regular limit?

When we have $\lim_{x \rightarrow a^{-}} f(x)$ and $\lim_{x \rightarrow a^{+}} f(x)$, we can also confirm if a regular limit or $\lim_{x \rightarrow a} f(x)$ exists.

Let’s say $\lim_{x \rightarrow a^{-}} f(x) = M$ and $\lim_{x \rightarrow a^{+}} f(x) = N$, the function $f(x)$ has a regular limit if and only if $M = N$.

### Summary of one sided limits definition and properties

Before we try out more problems that involve one-sided limits, why don’t we go ahead and recap everything that we’ve learned so far?

- When we’re looking for $\lim_{x \rightarrow a^{-}} f(x)$, this means that we want to see the values that $f(x)$ approaches from the left of $a$.
- Similarly, $\lim_{x \rightarrow a^{+}} f(x)$, means that we want to see the values that $f(x)$ approaches from the right of $a$.
- Regardless of what’s given to represent $f(x)$ (graphically, tabular, or its expression), always go back to the fundamental definition of one-sided limits.

*Example 1*

The graph of $f(x)$ is as shown below.

Determine the values of the following:

a. $\lim_{x \rightarrow 2^{-}} f(x)$

b. $\lim_{x \rightarrow 2^{+}} f(x)$

__Solution__

When working with one-sided limits, make sure that we’re working with the right intervals or side.

For $\lim_{x \rightarrow 2^{-}} f(x)$, we want to observe how $f(x)$ behaves based on the values to the left of $2$.

We can see that from the left, $x$ approaches $2$, the graph is approaching $4$.

a. This means that $\lim_{x \rightarrow 2^{-}} f(x) = 4$.

Inspecting the graph’s behavior from the right this time, we can see that as $x$ approaches $2$ from the right, the graph approaches $f(x) = 6$.

b. This means that $\lim_{x \rightarrow 2^{+}} f(x) = 6$.

*Example 2*

Describe the one-sided limits as $x$ approaches $4$ using the table of values for $f(x)$ as shown below.

$\boldsymbol{x}$ | $\boldsymbol{f(x)}$ | $\boldsymbol{x}$ | $\boldsymbol{f(x)}$ |

$3$ | $12.35789$ | $4.0001$ | $7.999999$ |

$3.5$ | $12.45687$ | $4.001$ | $7.894725$ |

$3.9$ | $12.65489$ | $4.01$ | $7.611480$ |

$3.99$ | $12.78954$ | $4.1$ | $7.421597$ |

$3.999$ | $12.96154$ | $4.5$ | $7.14057$ |

$3.9999$ | $12.99999$ | $5$ | $7.01245$ |

__Solution__

When given the table of values, divide the group to have a table that shows all values of $x$ that are less than $4$ and a group where $x$ are all greater than $4$.

Why don’t we inspect the limit from the left of $f(x)$ as $x$ approaches $4$?

$\boldsymbol{x}$ | $\boldsymbol{f(x)}$ |

$3$ | $12.35789$ |

$3.5$ | $12.45687$ |

$3.9$ | $12.65489$ |

$3.99$ | $12.78954$ |

$3.999$ | $12.96154$ |

$\color{blue} 3.9999$ | $ \color{blue} 12.99999$ |

We can see from the table that from the left, as $x$ approaches $4$, the value of $f(x)$ gets closer to $13$. This means that the value of $\lim_{x \rightarrow 4^{-}} f(x)$ is $13$.

We apply a similar approach but this time, we’re looking at the values of $x$ that are greater than $4$.

$\boldsymbol{x}$ | $\boldsymbol{f(x)}$ |

$\color{blue} 4.0001$ | $\color{blue} 7.999999$ |

$4.001$ | $7.894725$ |

$4.01$ | $7.611480$ |

$4.1$ | $7.421597$ |

$4.5$ | $7.14057$ |

$5$ | $7.01245$ |

We can see that as $x$ gets closer to $4$ from the right, the value of $f(x)$ approaches $8$. Hence, we have $\lim_{x \rightarrow 4^{+}} f(x) = 8$.

*Example 3*

The function $f(x)$ is a piecewise function as shown by the equation below.

$ f(x)=\left\{\begin{matrix}\dfrac{x}{x+1},& 0 \leq x < 1\\ \sqrt{25-x^2},&1 \leq x \leq 5\\4x – 5,&5 < x \leq 10\end{matrix}\right.$

Evaluate the following expressions:

a. $\lim_{x \rightarrow 1^{-}} f(x)$

b. $\lim_{x \rightarrow 1^{+}} f(x)$

c. $\lim_{x \rightarrow 3^{-}} f(x)$

d. $\lim_{x \rightarrow 3^{+}} f(x)$

Use the result from a to d to confirm if there is a regular limit for $f(x)$ as $x$ approaches $1$ and when $x$ approaches $2$.

__Solution__

To find the limit of $f(x)$ as $x \rightarrow 1^{-}$, we can find its limit by using the right expression for $f(x)$ at when $ x< 1$.

$f(x)=\left\{\begin{matrix}\color{blue}\dfrac{x}{x-1},& \color{blue}0 \leq x < 1\\ \sqrt{25-x^2},&1 \leq x \leq 5\\4x – 5,&5 < x \leq 10\end{matrix}\right.$

When $0 \leq x < 1$, $f(x) = \dfrac{x}{x + 1}$, so find the value of $f(1)$ using the expression of $f(x)$ at the given interval.

$\begin{aligned} \lim_{x \rightarrow 1^{-}} f(x) &= f(1), {\text{ where }} f(x) &= \dfrac{x}{x + 1}\\&=\dfrac{\color{blue} 1}{{\color{blue}1} + 1}\\&= \dfrac{1}{2}\end{aligned}$

a. We now have $\lim_{x \rightarrow 1^{-}} f(x) = \dfrac{1}{2}$.

Applying the same process for the limit from the right of $1$. This time though, we have $f(x) = \sqrt{25 – x^2}$ when $1 \leq x \leq 5$.

$\begin{aligned} \lim_{x \rightarrow 1^{+}} f(x) &= f(1), {\text{ where }} f(x) &= \sqrt{25 – x^2}\\&=\sqrt{25 -{\color{blue}1}^2 }\\&= \sqrt{24}\\&= 2\sqrt{6}\end{aligned}$

b. Hence, we have $\lim_{x \rightarrow 1^{+}} f(x) = 2\sqrt{6}$.

We will apply the same thought process for parts c and d.

Limit from left, $\boldsymbol{\lim_{x \rightarrow 3^{-}} f(x)}$ | Limit from right, $\boldsymbol{\lim_{x \rightarrow 3^{+}} f(x)}$ |

The values from the left of $3$ lies within $1 \leq x \leq 5$. $\begin{aligned} \lim_{x \rightarrow 3^{-}} f(x) &= f(3), {\text{ where }} f(x) &= \sqrt{25 – x^2}\\&=\sqrt{25 -{\color{blue}3}^2 }\\&= \sqrt{16}\\&= 4\end{aligned}$ | The values from the right of $3$ lies within $1 \leq x \leq 5$. $\begin{aligned} \lim_{x \rightarrow 3^{-}} f(x) &= f(3), {\text{ where }} f(x) &= \sqrt{25 – x^2}\\&=\sqrt{25 -{\color{blue}3}^2 }\\&= \sqrt{16}\\&= 4\end{aligned}$ |

c. This means that $\boldsymbol{$\lim_{x \rightarrow 3^{-}} f(x)} = 4$.

d. The second column also shows that $\boldsymbol{$\lim_{x \rightarrow 3^{+}} f(x)} = 4$.

Comparing the results of **a and b shows** that the **regular limit does not exist **since the two limits are not equal.

$\begin{aligned} \lim_{x \rightarrow 1^{-}} f(x) &= \dfrac{1}{2}\\\lim_{x \rightarrow 1^{+}} f(x) &= 2\sqrt{6}\\\lim_{x \rightarrow 1^{-}} f(x)&\neq \lim_{x \rightarrow 1^{+}} f(x)\end{aligned}$

Comparing the results of **c and d**, we can show that the **regular limit exists **since the two limits are equal.

$\begin{aligned} \lim_{x \rightarrow 3^{-}} f(x) &= 4\\\lim_{x \rightarrow 3^{+}} f(x) &= 4\\\lim_{x \rightarrow 3^{-}} f(x)&= \lim_{x \rightarrow 3^{+}} f(x)\end{aligned}$

### Practice Questions

*Images/mathematical drawings are created with GeoGebra.*