A.6 One-sided Limits

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:
\[f(x) =
\begin{cases}
-3 & \text{for } x \lt 0 \\
\phantom{-}5 & \text{for } x \ge 0
\end{cases}\] Graph of piecewise function f(x) to illustrate one-sided limits: The function equals -5 for x less than zero, and +5 for x greater than or equal to 0.
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.

“Limit from the left” and “limit from the right”

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:

One-sided Limits

• “–” means “from the left,” and is the left-hand limit, while
• “+” means “from the right” and is the right-hand limit.

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

Left-hand limit: $\lim_{x \to 0^{\bbox[yellow]{-}}} f(x) = -3,$ meaning “the limit as $x \to 0$ from the left is $-3.$”

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

Right-hand limit: $\lim_{x \to 0^{\bbox[yellow]{+}}} f(x) = 5,$ meaning “the limit as $x \to 0$ from the right is 5.”

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.

The general limit and one-sided limits at $x=a$

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:

For the limit to exist, the one-sided limits must be equal:
\[\lim_{x \to a}f(x) = L \text{ if and only if } \lim_{x \to a^-}f(x) = L \text{ and } \lim_{x \to a^+}f(x) = L\]

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!

The Upshot

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.
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 x.

Practice

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

Graph of piecewise function. Relevant part for this question: on the open interval (-2, 2) the graph is a horizontal line at y = -2. There is a closed circle at (2, 1), and then a horizontal line away from that point ending on an open circle at (3, 1).

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

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

Show/Hide Solution

Same graph, now with a red from the left pointing to (2, -2), and a red arrow from the right pointing to (2, 1). There is also a red star over the function's value at (2, 1).

The figure shows a red arrow on the left side of $x=2$, which indicates that the left-hand limit: $\displaystyle{ \lim_{x \to 2^{-} }f(x) }=-2. \; \blacktriangleleft $
Similarly, the red arrow on the right side of $x=2$ illustrates the right-hand limit: $\displaystyle{ \lim_{x \to 2^{+} }f(x) }=1. \; \blacktriangleleft $
Since the limit on the left-hand side of $x=2$ exists but is not equal to the limit on the right-hand side of $x=2$, then $\displaystyle{ \lim_{x \to 2 }f(x) }$ does not exist and is undefined. $ \; \blacktriangleleft $
The figure also shows a red star indicating the value of $y=f(x)$ at $x=2$ is $y=1. \; \blacktriangleleft $
Thus the correct choice is $ \text{ (C) } \; \cmark $

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Practice Problem #2

Graph of piecewise function. The relevant parts for this question: on the closed interval (-4, -2) the graph is an upward straight line toward (-2, 2.5), and there is a filled-in circle at that point. Then one the open interval (-2, 2) the graph is a horizontal line at y=-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

Show/Hide Solution

Same graph, now with a red arrow from the left pointing toward (-2, 2.5), and a red arrow from the right pointing to the point (-2, -2). There is also a red star over the function's value at (-2, 2.5).

The figure shows a red arrow on the left side of $x=-2$ which indicates that the left-hand limit: $\displaystyle{ \lim_{x \to -2^{-} }f(x) }=2.5. \; \blacktriangleleft $
Similarly, the red arrow on the right side of $x=-2$ illustrates the right-hand limit: $\displaystyle{ \lim_{x \to -2^{+} }f(x) }=-2. \; \blacktriangleleft $
Since the limit on the left-hand side of $x=-2$ exists but is not equal to the limit on the right-hand side of $x=-2$, then $\displaystyle{ \lim_{x \to -2 }f(x) }$ does not exist, and so is undefined. $ \; \blacktriangleleft $
The figure also shows with a red arrow the value of $y=f(x)$ at $x=-2$ is $y=2.5. \; \blacktriangleleft $
Thus the correct choice is $ \text{ (A) } \; \cmark $

[hide solution]

Practice Problem #3

Graph of piecewise function. Relevant part for this question: near $x=-4$ the quote-unquote curve is essentially a vee with its base at (-4, 0).

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

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

Show/Hide Solution

Same graph, now with red arrows pointing from the left and right toward the point (-4, 0). That point also has a red star on it.

The figure shows a red arrow on the left side of $x=-4$ which indicates that the left-hand limit: $\displaystyle{ \lim_{x \to -4^{-} }f(x) }=0.\; \blacktriangleleft $
Similarly, the red arrow on the right side of $x=-4$ illustrates the right-hand limit: $\displaystyle{ \lim_{x \to -4^{+} }f(x) }=0. \; \blacktriangleleft $
Since the limit on the left-hand side of $x=-4$ exists and is equal to the limit on the right-hand side of $x=-4:$ $\displaystyle{ \lim_{x \to -4 }f(x) }=0. \; \blacktriangleleft $
The figure also shows with a red star the value of $y=f(x)$ at $x=-4$ is $y=0. \; \blacktriangleleft $
Thus the correct choice is $ \text{ (D) } \; \cmark $

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Practice Problem #4

Graph of y=f(x), which shows essentially a hyperbola near x=0.

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}

Show/Hide Solution

Same graph, now with a red arrow indicating the steep downwardness of the left-curve as you approach x=0 from the left, and an upward pointing arrow indicating the steep upwardness of the right-curve as you approach x=0 from the right.

I. $\displaystyle{\lim_{x \to 0^-}f(x) = -\infty}$ is TRUE: the graph shows that $y \to \, – \infty$ as $x \to 0$ from the left.
II. $\displaystyle{\lim_{x \to 0^+}f(x) = \infty}$ is TRUE: the graph also shows $y \to + \infty$ as $x \to 0$ from the right.
III. $\displaystyle{\lim_{x \to 0}f(x) = \infty }$ is FALSE: this statement says that both the left- and right-hand limits are $+\infty,$ which is not true.
IV. $f(0) = \text{undefined}$ is TRUE: the function $f(x) = \dfrac{1}{x}$ is undefined at $x=0.$

Hence the true statements are I, II & IV $\implies \;\text{ (B)} \quad \cmark$

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