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)={−3for 𝑥<0−5for 𝑥≥0

As you'll remember, lim𝑥→0𝑓(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 <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 >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:

So for the function 𝑓(𝑥) above:

Left-hand limit: lim𝑥→0−𝑓(𝑥)=−3, meaning "the limit as 𝑥 →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 [indent]Right-hand limit: lim𝑥→0+𝑓(𝑥)=5, meaning "the limit as 𝑥 →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: Because lim𝑥→0−𝑓(𝑥)≠lim𝑥→0+𝑓(𝑥), lim𝑥→0𝑓(𝑥)=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:

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, lim𝑥→−2−𝑓(𝑥)=−3andlim𝑥→−2+𝑓(𝑥)=−3 and hence lim𝑥→−2𝑓(𝑥)=−3. And at 𝑥 =4, lim𝑥→4−𝑓(𝑥)=5andlim𝑥→4+𝑓(𝑥)=5 and hence lim𝑥→4𝑓(𝑥)=5.

Revisiting 𝑔(𝑥) =sin⁡(𝑥)𝑥

As a different example, let's consider another function we looked at earlier, 𝑔(𝑥)=sin⁡(𝑥)𝑥

For this function we have lim𝑥→0−𝑔(𝑥)=1andlim𝑥→0+𝑔(𝑥)=1 and hence lim𝑥→0𝑔(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!

Practice Problems

Question 1: Examining a piecewise function

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

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:
lim𝑥→𝑎−𝑓(𝑥),lim𝑥→𝑎+𝑓(𝑥),lim𝑥→𝑎𝑓(𝑥),and 𝑓(𝑎).

Answer Choice:	lim𝑥→𝑎−𝑓(𝑥)	lim𝑥→𝑎+𝑓(𝑥)	lim𝑥→𝑎𝑓(𝑥)	𝑓(𝑎)
(A)	1	-2	Undefined	-1
(B)	1	-2	0	Undefined
(C)	-2	1	Undefined	1
(D)	-2	1	Undefined	Undefined

✓✘A✓✘B✓✘C✓✘D

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The figure shows a red arrow on the left side of 𝑥 =2, which indicates that the left-hand limit: lim𝑥→2−𝑓(𝑥) = −2. ◂
Similarly, the red arrow on the right side of 𝑥 =2 illustrates the right-hand limit: lim𝑥→2+𝑓(𝑥) =1. ◂
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 lim𝑥→2𝑓(𝑥) does not exist and is undefined. ◂
The figure also shows a red star indicating the value of 𝑦 =𝑓(𝑥) at 𝑥 =2 is 𝑦 =1. ◂
Thus the correct choice is  (C)  ✓

Practice Problem 2

Given 𝑦 =𝑓(𝑥) for the interval [ −5, 4] is shown, find the following values at 𝑎 = −2: lim𝑥→𝑎−𝑓(𝑥),lim𝑥→𝑎+𝑓(𝑥),lim𝑥→𝑎𝑓(𝑥),and 𝑓(𝑎).

Answer Choice:	lim𝑥→𝑎−𝑓(𝑥)	lim𝑥→𝑎+𝑓(𝑥)	lim𝑥→𝑎𝑓(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

✓✘A✓✘B✓✘C✓✘D

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The figure shows a red arrow on the left side of 𝑥 = −2 which indicates that the left-hand limit: lim𝑥→−2−𝑓(𝑥) =2.5. ◂
Similarly, the red arrow on the right side of 𝑥 = −2 illustrates the right-hand limit: lim𝑥→−2+𝑓(𝑥) = −2. ◂
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 lim𝑥→−2𝑓(𝑥) does not exist, and so is undefined. ◂
The figure also shows with a red arrow the value of 𝑦 =𝑓(𝑥) at 𝑥 = −2 is 𝑦 =2.5. ◂
Thus the correct choice is  (A)  ✓

Practice Problem 3

Given 𝑦 =𝑓(𝑥) for the interval [ −5, 4] is shown, find the following values at 𝑎 = −4:
lim𝑥→𝑎−𝑓(𝑥),lim𝑥→𝑎+𝑓(𝑥),lim𝑥→𝑎𝑓(𝑥),and 𝑓(𝑎).

Answer Choice:	lim𝑥→𝑎−𝑓(𝑥)	lim𝑥→𝑎+𝑓(𝑥)	lim𝑥→𝑎𝑓(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

✓✘A✓✘B✓✘C✓✘D

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The figure shows a red arrow on the left side of 𝑥 = −4 which indicates that the left-hand limit: lim𝑥→−4−𝑓(𝑥) =0. ◂
Similarly, the red arrow on the right side of 𝑥 = −4 illustrates the right-hand limit: lim𝑥→−4+𝑓(𝑥) =0. ◂
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 : lim𝑥→−4𝑓(𝑥) =0. ◂
The figure also shows with a red star the value of 𝑦 =𝑓(𝑥) at 𝑥 = −4 is 𝑦 =0. ◂
Thus the correct choice is  (D)  ✓

Practice Problem 4

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

I. lim𝑥→0−𝑓(𝑥)=−∞
II. lim𝑥→0+𝑓(𝑥)=∞
III. lim𝑥→0𝑓(𝑥)=∞
IV. 𝑓(0) =undefined

(A) I, II, III, and IV(B) I, II, and IV(C) IV only (D) I and II(E) III and IV

✓✘A✓✘B✓✘C✓✘D✓✘E

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I. lim𝑥→0−𝑓(𝑥)=−∞ is TRUE: the graph shows that 𝑦 → −∞ as 𝑥 →0 from the left.
II. lim𝑥→0+𝑓(𝑥)=∞ is TRUE: the graph also shows 𝑦 → +∞ as 𝑥 →0 from the right.
III. lim𝑥→0𝑓(𝑥)=∞ is FALSE: this statement says that both the left- and right-hand limits are +∞, which is not true.
IV. 𝑓(0) =undefined is TRUE: the function 𝑓(𝑥) =1𝑥 is undefined at 𝑥 =0.

Hence the true statements are I, II & IV ⟹  (B) ✓

On the next screen, we'll consider some functions that grow without bound, heading to ∞ or −∞ at a particular value of x.

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

The Upshot

1. 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.
2. For the limit at a point to exist, the one-sided limits at that point must be equal: lim𝑥→𝑎𝑓(𝑥)=𝐿 if and only if lim𝑥→𝑎−𝑓(𝑥)=𝐿 and lim𝑥→𝑎+𝑓(𝑥)=𝐿