okay, so I _think_ I'm starting to understand, is a limit a value that we would expect there to be, if the existing trend continued, regardless of what the actual value is?

Yes certainly it aims on what the trend is and not on what the function was. It tells what the trend approaches as we go to a particular point in a graph.

how did you determine problem 4? i dont understand the correlation between the limit and it strictly increasing

Correct me if i'm wrong but as the function has a discontinuity at x=5 and the closest values we have to lim->x=5 is at x=4 and x=6, it made sense to me that with the given information the most accurate estimation of lim->x=5 would be the average value of x=4 and x=6. I.e (4.9+5.6)/2 which is 5.25 which is approximately 5.3

So in problem 3, if x=7 is not an asymptote, what is it called?

If x=7 was an asymptote, we would see that as it approaches 7 from one side, it would start to approach infinity at a dramatic pace and do the same or negative infinity from the other. As you can see, they both approach −1.925 from each side. Therefore we have discontinuity which is when a point at which a function is discontinuous or undefined.

I dont agree with the answer... Obviously It can be true... but not only this one can be true. Let's show why... Imagine those values. f(3) = 4,3 (as shown) f(4) = 4,9 (as shown) f(4,5) = 5,2 f(4,99) = 5,3 f(5) = 4,8 (as shown) f(5,001) = 4,9 f(5,1) = 5 f(5,5) = 5,5 f(6) = 5,6 (as shown) Which acomplish our requirements... it is increasing everywhere (understood as positive derivative)... And in this case there is no limit. The approximations from both sides is different (has different value)

That's why the problem asks for a 'reasonable estimate' of the limit. We're given incomplete information, so we can only make an educated guess of the limit if we assume the function behaves nicely.

the function was increasing by six before and after x=5 but when x=5 the function becomes messy, because the value 5 is out of place. So why is this still the right answer?

are u talking about the last question? it says the function never decreases except at the point x=5. This means there must be a point discontinuity. to find the limit as x approaches 5, we have to do some guessing. at x=4, f(x)=4.9 while at x=6, f(x)=5.6. Thus, we know that the limit value must be between 4.9 and 5.6. The only value that falls in between that range is 5.3 and thus that is the right answer. hope this helps

An asymptote is an x-value (in some cases even y-value) that a function gets extremely close to but never touches. It is graphed as a dashed line. If x = 0 is our vertical asymptote of our function f, then the limit notation is- The Limit as x approaches infinity of f(x) = 0

I am confused about question 4, even after the tutor explained it i am still having trouble. Can someone help me?

In problem 4, it is said that "The function is increasing everywhere except at x=5", which means that x=5 most certainly is a function jump. That said, x=5 is presumably a single dot out of the rest of the graphic of f, so the limit of f(x) when x approaches 5 is different from f(5). However, knowing that the limit depends on the points surrounding it, and knowing that *f(4) < that limit < f(6), i.e 4.9 < that limit < 5.6*, we can deduced that a good estimate for the limit in question is 5.3. Hope this helps!

What is the difference between the function value and the limit value?

The function value at a point a refers to the value f(a) whereas the limit value refers to the limit (f(x)) as x approaches a. Both values are the same for a continuous function but might be different for a function discontinuous at a.

Main content

AP®︎/College Calculus BC

Course: AP®︎/College Calculus BC > Unit 1

Lesson 4: Estimating limit values from tables

Using tables to approximate limit values

Q: What is the difference between the function value and the limit value?

The function value at a point a refers to the value f(a) whereas the limit value refers to the limit (f(x)) as x approaches a. Both values are the same for a continuous function but might be different for a function discontinuous at a.

Google Classroom

Tables can be a powerful tool to approximate a limit, but they need to be used wisely. Learn how to create tables in order to find a good approximation of a limit, and learn how to approximate a limit given a table of values.

Limits are a tool for reasoning about function behavior, and tables are a tool for reasoning about limits. One nice thing about tables is that we can get more precise estimates of limits than we'd get by eyeballing graphs.

When using a table to approximate limits, it's important to create it in a way that simulates the feeling of getting "infinitely close" to some desired

x

-value.

Example

Imagine we're asked to approximate this limit:

lim_{x \to 2} \frac{x - 2}{x^{2} - 4}

Note: The function is actually undefined at

x = 2

because the denominator evaluates to zero, but the limit as

x

approaches

2

still exists.

Step 1: We'd like to pick a value that's a little bit less than

x = 2

(that is, a value that's "to the left" of

2

when thinking about the standard

x

-axis), so maybe start with something like

x = 1.9

$x$ ‍	$1.9$ ‍	$2$ ‍
$f (x)$ ‍	$0.2564$ ‍	undefined

Step 2: Try a couple more

x

-values to simulate the feeling of getting infinitely close to

x = 2

from the left.

$x$ ‍	$1.9$ ‍	$1.99$ ‍	$1.9999$ ‍	$2$ ‍
$f (x)$ ‍	$0.2564$ ‍	$0.2506$ ‍	$0.25001$ ‍	undefined

Notice how our

x

-values

{1.9, 1.99, 1.9999}

really "zoom in" around

x = 2

. A worse choice of

x

-values would have been constant increments like

{- 1, 0, 1}

, which aren't very helpful for thinking about getting infinitely close to

x = 2

Step 3: Approach

x = 2

from the right just like we did from the left. We want to do this in a way that simulates the feeling of getting infinitely close to

x = 2

$x$ ‍	$1.9$ ‍	$1.99$ ‍	$1.9999$ ‍	$2.0001$ ‍	$2.01$ ‍	$2.1$ ‍
$f (x)$ ‍	$0.2564$ ‍	$0.2506$ ‍	$0.25001$ ‍	$0.24999$ ‍	$0.2494$ ‍	$0.2439$ ‍

(Note: We've removed

x = 2

from the table to save space, and also because it isn't necessary for reasoning about the limit value.)

Looking at the table we've created, we have very strong evidence that the limit is

0.25

. But, if we're honest with ourselves, we must admit that what we have is only a reasonable approximation. We can't say for sure that this is the actual value of the limit.

Problem 1

Three students were given a function

f

and asked to estimate

lim_{x \to 2} f (x)

. Each student created a table (shown below).

Each table is accurate, but which one is the best for approximating the limit?

(Choice A)
A $x$ ‍ $- 1$ ‍ $0$ ‍ $1$ ‍ $2$ ‍ $3$ ‍ $4$ ‍
$f (x)$ ‍ $0.2$ ‍ $1.3$ ‍ $2.9$ ‍ $3.25$ ‍ $2.9$ ‍ $1.3$ ‍
(Choice B)
B $x$ ‍ $1.9$ ‍ $1.99$ ‍ $1.999$ ‍ $2.001$ ‍ $2.01$ ‍ $2.1$ ‍
$f (x)$ ‍ $3.32$ ‍ $3.264$ ‍ $3.251$ ‍ $3.249$ ‍ $3.242$ ‍ $3.31$ ‍
(Choice C)
C $x$ ‍ $2.001$ ‍ $2.01$ ‍ $2.1$ ‍ $2.25$ ‍ $2.5$ ‍ $3$ ‍
$f (x)$ ‍ $3.249$ ‍ $3.242$ ‍ $3.31$ ‍ $3.2$ ‍ $3.05$ ‍ $2.9$ ‍

A	$x$ ‍	$- 1$ ‍	$0$ ‍	$1$ ‍	$2$ ‍	$3$ ‍	$4$ ‍
	$f (x)$ ‍	$0.2$ ‍	$1.3$ ‍	$2.9$ ‍	$3.25$ ‍	$2.9$ ‍	$1.3$ ‍

A	$x$ ‍	$- 1$ ‍	$0$ ‍	$1$ ‍	$2$ ‍	$3$ ‍	$4$ ‍
	$f (x)$ ‍	$0.2$ ‍	$1.3$ ‍	$2.9$ ‍	$3.25$ ‍	$2.9$ ‍	$1.3$ ‍

B	$x$ ‍	$1.9$ ‍	$1.99$ ‍	$1.999$ ‍	$2.001$ ‍	$2.01$ ‍	$2.1$ ‍
	$f (x)$ ‍	$3.32$ ‍	$3.264$ ‍	$3.251$ ‍	$3.249$ ‍	$3.242$ ‍	$3.31$ ‍

B	$x$ ‍	$1.9$ ‍	$1.99$ ‍	$1.999$ ‍	$2.001$ ‍	$2.01$ ‍	$2.1$ ‍
	$f (x)$ ‍	$3.32$ ‍	$3.264$ ‍	$3.251$ ‍	$3.249$ ‍	$3.242$ ‍	$3.31$ ‍

C	$x$ ‍	$2.001$ ‍	$2.01$ ‍	$2.1$ ‍	$2.25$ ‍	$2.5$ ‍	$3$ ‍
	$f (x)$ ‍	$3.249$ ‍	$3.242$ ‍	$3.31$ ‍	$3.2$ ‍	$3.05$ ‍	$2.9$ ‍

C	$x$ ‍	$2.001$ ‍	$2.01$ ‍	$2.1$ ‍	$2.25$ ‍	$2.5$ ‍	$3$ ‍
	$f (x)$ ‍	$3.249$ ‍	$3.242$ ‍	$3.31$ ‍	$3.2$ ‍	$3.05$ ‍	$2.9$ ‍

Want more practice? Try this exercise.

Common mistakes when creating tables to estimate limits

Here are a several things to watch out for as you create your own tables to approximate limits:

Assuming the function value is the limit value: The example above highlights a case where the function is undefined, yet the limit still exists. Avoid jumping to conclusions about the limit value based on the function value.

Not getting infinitely close: Getting infinitely close means we’re trying to get so close to a desired

x

-value that there’s very little room left between where we are and where that value is—close enough to convince us that the estimate we’re getting is most likely what the limit is.

Avoid picking $x$ ‍-values in constant increments like ${- 1, 0, 1}$ ‍ or even ${1.91, 1.92, 1.93}$ ‍ because those values don't really get us infinitely close—they only get us kind of close. To get infinitely close, we want to keep reducing our increments, using $x$ ‍-values like ${1.9, 1.99, 1.999}$ ‍, so that we’re shrinking the space between where we are and where we’d like to be.

Not approaching from both sides: Remember to approach your desired

x

-value from both the left and the right. Remember, for the limit to exist, the left- and right-hand limits must be equal. Avoid jumping to conclusions about the limit value after only approaching your desired $x$ ‍-value from one side.

Assuming "left side" means "negative": Some students mistakenly believe that when approaching from the left they must use negative numbers. In the example above, we approached

x = 2

from the left by using positive values that were just a little bit less than

2

, such as

1.9

and

1.99

. Don't assume you must use negative $x$ ‍-values when approaching from the left.

Problem 2

The function

g

is defined over the real numbers. This table gives select values of

g

$x$ ‍	$g (x)$ ‍
$4$ ‍	$3.37$ ‍
$4.9$ ‍	$3.5$ ‍
$4.99$ ‍	$3.66$ ‍
$4.999$ ‍	$3.68$ ‍
$5$ ‍	$6.37$ ‍
$5.001$ ‍	$3.68$ ‍
$5.01$ ‍	$3.7$ ‍
$5.1$ ‍	$3.84$ ‍
$6$ ‍	$3.97$ ‍

What is a reasonable estimate for $lim_{x \to 5} g (x)$ ‍?

Want more practice? Try this exercise.

Common mistakes when estimating limits from tables

Confusing the limit value with the function value: Remember that the limit of a function at a certain point isn't necessarily equal to the function's value at that point. For example, in Problem 2,

g (5) = 6.37

but

lim_{x \to 5} g (x)

is about

3.68

Thinking a limit value is always an integer: Some limits are "nice" and have integer values or nice fraction values. For example, the limit in our first example here was

0.25

. Some limits are less nice, like the limit in Problem 2 which is somewhere around

3.68

Summary questions

Problem 3

A student created a table to help them reason about

lim_{x \to 7} g (x)

$x$ ‍	$6$ ‍	$6.99$ ‍	$6.9999$ ‍	$7$ ‍	$7.0001$ ‍	$7.01$ ‍	$8$ ‍
$g (x)$ ‍	$- 3.41$ ‍	$- 1.94$ ‍	$- 1.9252$ ‍	undefined	$- 1.9248$ ‍	$- 1.91$ ‍	$0.46$ ‍

Based on the table, what can you reasonably conclude about the limit?

(Choice A)
The value $- 1.925$ ‍ is a reasonable estimate of $lim_{x \to 7} g (x)$ ‍.
(Choice B)
There appears to be an asymptote at $x = 7$ ‍.
(Choice C)
The limit doesn't exist because the function is undefined at $x = 7$ ‍.
(Choice D)
As we approach $x = 7$ ‍, the value of $g (x)$ ‍ definitely approaches $- 1.925$ ‍ exactly.

Problem 4

The table gives a few values of function

f

. The function is increasing everywhere except at

x = 5

, and

lim_{x \to 5} f (x)

exists.

$x$ ‍	$2$ ‍	$3$ ‍	$4$ ‍	$5$ ‍	$6$ ‍	$7$ ‍	$8$ ‍
$f (x)$ ‍	$3.7$ ‍	$4.3$ ‍	$4.9$ ‍	$4.8$ ‍	$5.6$ ‍	$6.2$ ‍	$6.9$ ‍

Which is a reasonable estimate for $lim_{x \to 5} f (x)$ ‍?

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Nileghi
Posted 6 years ago. Direct link to Nileghi's post “So in problem 3, if x=7 i...”
So in problem 3, if x=7 is not an asymptote, what is it called?
Button navigates to signup pageComment on Nileghi's post “So in problem 3, if x=7 i...”
(12 votes)
Answer
- Skift02
  Posted 6 years ago. Direct link to Skift02's post “If x=7 was an asymptote, ...”
  If x=7 was an asymptote, we would see that as it approaches 7 from one side, it would start to approach infinity at a dramatic pace and do the same or negative infinity from the other. As you can see, they both approach −1.925 from each side. Therefore we have discontinuity which is when a point at which a function is discontinuous or undefined.
  Comment on Skift02's post “If x=7 was an asymptote, ...”
  (90 votes)
Dahlia Williams
Posted 6 years ago. Direct link to Dahlia Williams's post “how did you determine pro...”
how did you determine problem 4? i dont understand the correlation between the limit and it strictly increasing
Button navigates to signup pageComment on Dahlia Williams's post “how did you determine pro...”
(21 votes)
Answer
- David.CM.Bergman
  Posted 5 years ago. Direct link to David.CM.Bergman's post “Correct me if i'm wrong b...”
  Correct me if i'm wrong but as the function has a discontinuity at x=5 and the closest values we have to lim->x=5 is at x=4 and x=6, it made sense to me that with the given information the most accurate estimation of lim->x=5 would be the average value of x=4 and x=6. I.e (4.9+5.6)/2 which is 5.25 which is approximately 5.3
  Comment on David.CM.Bergman's post “Correct me if i'm wrong b...”
  (21 votes)
Elijah Groves
Posted 5 years ago. Direct link to Elijah Groves's post “okay, so I _think_ I'm st...”
okay, so I think I'm starting to understand, is a limit a value that we would expect there to be, if the existing trend continued, regardless of what the actual value is?
Button navigates to signup pageButton navigates to signup page
(22 votes)
Answer
- apasa2k
  Posted 2 years ago. Direct link to apasa2k's post “Yes certainly it aims on ...”
  Yes certainly it aims on what the trend is and not on what the function was. It tells what the trend approaches as we go to a particular point in a graph.
  Button navigates to signup page
  (3 votes)
Jordi Manyà
Posted 4 years ago. Direct link to Jordi Manyà's post “I dont agree with the ans...”
I dont agree with the answer... Obviously It can be true... but not only this one can be true. Let's show why... Imagine those values.
f(3) = 4,3 (as shown)
f(4) = 4,9 (as shown)
f(4,5) = 5,2
f(4,99) = 5,3
f(5) = 4,8 (as shown)
f(5,001) = 4,9
f(5,1) = 5
f(5,5) = 5,5
f(6) = 5,6 (as shown)
Which acomplish our requirements... it is increasing everywhere (understood as positive derivative)... And in this case there is no limit. The approximations from both sides is different (has different value)
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- kubleeka
  Posted 4 years ago. Direct link to kubleeka's post “That's why the problem as...”
  That's why the problem asks for a 'reasonable estimate' of the limit. We're given incomplete information, so we can only make an educated guess of the limit if we assume the function behaves nicely.
  Button navigates to signup page
  (13 votes)
CaptAngryEyes
Posted 6 months ago. Direct link to CaptAngryEyes's post “On the last problem shoul...”
On the last problem shouldn't it be specified that the function isn't continuous at 5?
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(10 votes)
Answer
353792632
Posted 5 months ago. Direct link to 353792632's post “the function was increasi...”
the function was increasing by six before and after x=5 but when x=5 the function becomes messy, because the value 5 is out of place. So why is this still the right answer?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Forever Learner
  Posted 5 months ago. Direct link to Forever Learner's post “are u talking about the l...”
  are u talking about the last question? it says the function never decreases except at the point x=5. This means there must be a point discontinuity. to find the limit as x approaches 5, we have to do some guessing. at x=4, f(x)=4.9 while at x=6, f(x)=5.6. Thus, we know that the limit value must be between 4.9 and 5.6. The only value that falls in between that range is 5.3 and thus that is the right answer. hope this helps
  Comment on Forever Learner's post “are u talking about the l...”
  (11 votes)
lauren
Posted 6 years ago. Direct link to lauren's post “I am confused about quest...”
I am confused about question 4, even after the tutor explained it i am still having trouble. Can someone help me?
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(2 votes)
Answer
- Joana
  Posted 5 years ago. Direct link to Joana's post “In problem 4, it is said ...”
  In problem 4, it is said that "The function is increasing everywhere except at x=5", which means that x=5 most certainly is a function jump. That said, x=5 is presumably a single dot out of the rest of the graphic of f, so the limit of f(x) when x approaches 5 is different from f(5).
  However, knowing that the limit depends on the points surrounding it, and knowing that f(4) < that limit < f(6), i.e 4.9 < that limit < 5.6, we can deduced that a good estimate for the limit in question is 5.3.
  Hope this helps!
  Comment on Joana's post “In problem 4, it is said ...”
  (12 votes)
Ciara Turner
Posted 6 years ago. Direct link to Ciara Turner's post “What is the difference be...”
What is the difference between the function value and the limit value?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Hemang Rajvanshy
  Posted 6 years ago. Direct link to Hemang Rajvanshy's post “The function value at a p...”
  The function value at a point a refers to the value f(a) whereas the limit value refers to the limit (f(x)) as x approaches a.
  Both values are the same for a continuous function but might be different for a function discontinuous at a.
  Comment on Hemang Rajvanshy's post “The function value at a p...”
  (9 votes)
oe
Posted 8 months ago. Direct link to oe's post “On this post (Lesson 4: E...”
On this post (Lesson 4: Estimating limit values from tables of the COURSE: AP College Calculus BC, Unit 1) Summary Questions, Problem 4, It has a flaw.

Since it said: "The function is increasing everywhere except at x=5"
It means the f(x) is not increasing or keep the same at the next point slightly larger than the one at x=5; or the f(x) at x=5 simply not exist, so that there won't be a f(x) to increase or not.
However, as the answer indicates that "the reasonable estimate of lim x->5 f(x) is 5.3", it means the function of the next point slightly larger than the one of x=5 is 5.3, which means from the function of x=5, it increased, which is inappropriate to what indicates in the question.
Howbeit, it may means that from the side of x=5-, the function of the x slightly smaller than 5, which has a limit to 5.3, dropped to f(5), 4.8.
Nevertheless, from f(5)=4.8 to the function of the side of 5+, it increased.

I hope the content specialist think about this angle and update the problem set soon. Thank you!
Button navigates to signup pageComment on oe's post “On this post (Lesson 4: E...”
(5 votes)
Answer
huzaifa abedeen
Posted 2 years ago. Direct link to huzaifa abedeen's post “What is asymptote?”
What is asymptote?
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(4 votes)
Answer
- Ash_001
  Posted 2 years ago. Direct link to Ash_001's post “An asymptote is an x-valu...”
  An asymptote is an x-value (in some cases even y-value) that a function gets extremely close to but never touches. It is graphed as a dashed line.
  If x = 0 is our vertical asymptote of our function f, then the limit notation is-
  The Limit as x approaches infinity of f(x) = 0
  Button navigates to signup page
  (3 votes)