On this screen we consider one-sided limits, of course including practice problems for you to try with complete solutions available — all for free to support your learning.

Revisiting the function that “jumps” from the preceding screen

Recall one of the functions we considered on the previous screen for which the limit does *not* exist:

\[f(x) =

\begin{cases}

-3 & \text{for } x \lt 0 \\

\phantom{-}5 & \text{for } x \ge 0

\end{cases}\]

As you’ll remember, $\displaystyle{\lim_{x \to 0} f(x) }$ does *not* exist because the function “jumps” at $x=0$ from $-5$ to 5, and so there is no single number *L* that we can be as close to as we’d like for values of *x* both less than *and* greater than $x=0.$ That is, there is no single place on the graph you can point to that represents the limit at $x=0.$

You probably have the thought, though, that if we could consider only values of $x \lt 0,$ and remain always on the left-side of the vertical axis, then the limit *does* exist, and equals $-3$. If so, you’re right! Similarly, if we consider values of $x \gt 0$ and stay to the right of the vertical axis, then the limit again exists, and equals 5.

By examining the behavior of a function on either one side or the other of a point of interest, we are considering **one-sided limits**. The only real thing to remember is what the notation means:

• “–” means “*from* the left,” and is the **left-hand limit**, while

• “+” means “*from* the right” and is the **right-hand limit**.

• “+” means “

So for the function $f(x)$ above:

That is, we can get as close to the function’s output value of $-3$ as we’d like

by being sufficiently close to $x=0$ *from the left.*

And

That is, we can get as close to the function’s output value of 5 as we’d like

by being sufficiently close to $x=0$ *from the right.*

Still thinking about the function above, we can now write (very mathily!) using the one-sided limit notation:

\[\text{Because } \lim_{x \to 0^-}f(x) \ne \lim_{x \to 0+}f(x) \text{, } \lim_{x \to 0}f(x) = \text{DNE}\]
In words: because the limit from the left does not equal the limit from the right, the (general) limit does not exist at $x=0.$ (We include the word “general” there in parentheses only to indicate that we do not mean either one-sided limit. Typically you wouldn’t write that word; instead the word “limit” alone means you have to consider what happens as you approach the point of interest from both directions.)

Indeed, it is always true that the (general) limit exists and equals *L* only if the limit from both sides equals the same value *L*:

\[\lim_{x \to a}f(x) = L \text{ if and only if } \lim_{x \to a^-}f(x) = L \text{ and } \lim_{x \to a^+}f(x) = L\]

Continuing to think about our example piecewise function above, $f(x),$ if we consider any point *other* than $x=0,$ then the (general) limit exists because the one-sided limits are equal.

For instance, at $x=-2,$

\[\lim_{x \to -2^-}f(x) = -3 \quad \text{and} \quad \lim_{x \to -2+}f(x)= -3 \]
\[\text{and hence } \lim_{x \to -2}f(x) = -3.\]
And at $x= 4,$

\[\lim_{x \to 4^-}f(x) =5 \quad \text{and} \quad \lim_{x \to 4^+}f(x)= 5 \]
\[\text{and hence } \lim_{x \to 4}f(x) = 5.\]

Revisiting $g(x) = \dfrac{\sin(x)}{x}$

As a different example, let’s consider another function we looked at earlier,

\[g(x) = \dfrac{\sin(x)}{x}\]

For this function we have

\[\lim_{x \to 0^-}g(x) = 1 \quad \text{and} \quad \lim_{x \to 0^+}g(x) = 1 \]
\[\text{and hence } \lim_{x \to 0}g(x) = 1.\]
This is of course the same conclusion we reached earlier; we’re merely using the function to illustrate that when the one-sided limits are the same, the (general) limit exists, and equals the value of both one-sided limits.

Nothing more to discuss here; practice problems for you to try are below!

Questions or comments about anything on this screen? Please let us know on the Forum!

- When writing a one-sided limit,
- “–” means “
*from*the left,” and is the**left-hand limit**, while - “+” means “
*from*the right” and is the**right-hand limit**.

- “–” means “
- For the limit at a point to exist, the one-sided limits at that point must be equal:

\[\lim_{x \to a}f(x) = L \text{ if and only if } \lim_{x \to a^-}f(x) = L \text{ and } \lim_{x \to a^+}f(x) = L\]

On the next screen, we’ll consider some functions that grow without bound, heading to $\infty$ or $-\infty$ at a particular value of

Question 1 is typical of homework and AP^{®} multiple-choice type questions.

Question 1: Examining a piecewise function

The graph below shows the function $f(x)$ for the interval $[-5, 4].$

Use the graph to find the following values:

The first Practice Problem below combines all of the parts of the preceding Question into one answer choice, using a single row of the table shown to display the different quantities. Your job is to choose the row [(a), (b), (c), …] with the correct values. While this first problem follows directly from the Question above, the problems after that will simply ask you to determine each correct value and then the correct answer row.

Practice Problem #1

Given $y=f(x)$ for the interval $[-5,~4]$ is shown, find the following values at $a=2$.

\[\lim_{x \to a^{-}} f(x), \; \lim_{x \to a^{+}} f(x), \; \lim_{x \to a} f(x), \; \text{and } f(a). \]

\[\lim_{x \to a^{-}} f(x), \; \lim_{x \to a^{+}} f(x), \; \lim_{x \to a} f(x), \; \text{and } f(a). \]

Answer Choice: | $\displaystyle{ \lim_{x \to a^{-}} f(x)}$ | $\displaystyle{ \lim_{x \to a^{+}} f(x)}$ | $\displaystyle{ \lim_{x \to a} f(x)}$ | $f(a)$ |
---|---|---|---|---|

(A) | 1 | -2 | Undefined | -1 |

(B) | 1 | -2 | 0 | Undefined |

(C) | -2 | 1 | Undefined | 1 |

(D) | -2 | 1 | Undefined | Undefined |

Practice Problem #2

Given $y=f(x)$ for the interval $[-5,~4]$ is shown, find the following values at $a=-2$.
\[\lim_{x \to a^{-}} f(x), \; \lim_{x \to a^{+}} f(x), \; \lim_{x \to a} f(x), \; \text{and } f(a). \]

Answer Choice: | $\displaystyle{ \lim_{x \to a^{-}} f(x)}$ | $\displaystyle{ \lim_{x \to a^{+}} f(x)}$ | $\displaystyle{ \lim_{x \to a} f(x)}$ | $f(a)$ |
---|---|---|---|---|

(A) | 2.5 | -2 | Undefined | 2.5 |

(B) | -2 | 2.5 | 2.5 | -2 |

(C) | 2.5 | -2 | Undefined | Undefined |

(D) | -2 | 2.5 | Undefined | Undefined |

Practice Problem #3

Given $y=f(x)$ for the interval $[-5,~4]$ is shown, find the following values at $a=-4$.

\[\lim_{x \to a^{-}} f(x), \; \lim_{x \to a^{+}} f(x), \; \lim_{x \to a} f(x), \; \text{and } f(a). \]

\[\lim_{x \to a^{-}} f(x), \; \lim_{x \to a^{+}} f(x), \; \lim_{x \to a} f(x), \; \text{and } f(a). \]

Answer Choice: | $\displaystyle{ \lim_{x \to a^{-}} f(x)}$ | $\displaystyle{ \lim_{x \to a^{+}} f(x)}$ | $\displaystyle{ \lim_{x \to a} f(x)}$ | $f(a)$ |
---|---|---|---|---|

(A) | 0 | 0 | 0 | Undefined |

(B) | -5 | -2 | Undefined | Undefined |

(C) | 1.5 | 2.5 | Undefined | 0 |

(D) | 0 | 0 | 0 | 0 |

Practice Problem #4

Consider the function $f(x) = \dfrac{1}{x},$ graphed above. Which of the following statements are true?

I. $\displaystyle{\lim_{x \to 0^-}f(x) = -\infty }$

II. $\displaystyle{\lim_{x \to 0^+}f(x) = \infty }$

III. $\displaystyle{\lim_{x \to 0}f(x) = \infty }$

IV. $f(0) = \text{undefined}$

\begin{array}{lllll} \text{(A) I, II, III, & IV} && \text{(B) I, II, & IV} && \text{(C) IV only} && \text{(D) I & II} && \text{(E) III & IV} \end{array}II. $\displaystyle{\lim_{x \to 0^+}f(x) = \infty }$

III. $\displaystyle{\lim_{x \to 0}f(x) = \infty }$

IV. $f(0) = \text{undefined}$