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Finding Limits: Numerical and Graphical Approaches

Finding Limits: Numerical and Graphical Approaches

By the end of this section, you will be able to:

  • Understand limit notation
  • Find a limit using a graph
  • Find a limit using a table

Intuitively, we know what a limit is. A car can go only so fast and no faster. A trash can might hold 3333 gallons and no more. It is natural for measured amounts to have limits. What, for instance, is the limit to the height of a woman? The tallest woman on record was Jinlian Zeng from China, who was 8 ft 1 in. Is this the limit of the height to which women can grow? Perhaps not, but there is likely a limit that we might describe in inches if we were able to determine what it was.

To put it mathematically, the function whose input is a woman and whose output is a measured height in inches has a limit. In this section, we will examine numerical and graphical approaches to identifying limits.

Understanding Limit Notation

We have seen how a sequence can have a limit, a value that the sequence of terms moves toward as the number of terms increases. For example, the terms of the sequence

1,12,14,18,1,\tfrac{1}{2},\tfrac{1}{4},\tfrac{1}{8},\ldots

get closer and closer to 00. A sequence is one type of function, but functions that are not sequences can also have limits. We can describe the behavior of the function as the input values get close to a specific value. If the limit of a function f(x)=L,f(x)=L\text{,} then as the input xx gets closer and closer to a,a, the output yy-coordinate gets closer and closer to L.L. We say that the output “approaches” L.L.

The graph below provides a visual representation of the mathematical concept of limit. As the input value xx approaches a,a, the output value f(x)f(x) approaches L.L.

We write the equation of a limit as

limxaf(x)=L.\lim\limits_{x\to a}f(x)=L.

This notation indicates that as xx approaches aa both from the left of x=ax=a and the right of x=a,x=a, the output value approaches L.L.

Consider the function

f(x)=x26x7x7.f(x)=\tfrac{x^2-6x-7}{x-7}.

We can factor the function as shown.

f(x)=(x7)(x+1)x7Cancel like factors in numerator and denominator.f(x)=x+1, x7Simplify. \begin{array}{ll} f(x)=\tfrac{(x-7)(x+1)}{x-7} & \text{Cancel like factors in numerator and denominator.} \\[4pt] f(x)=x+1,\ x\ne7 & \text{Simplify.} \end{array}

Notice that xx cannot be 7,7, or we would be dividing by 0,0, so 77 is not in the domain of the original function. In order to avoid changing the function when we simplify, we set the same condition, x7,x\ne7, for the simplified function. We can represent the function graphically as shown below.

What happens at x=7x=7 is completely different from what happens at points close to x=7x=7 on either side. The notation

limx7f(x)=8\lim\limits_{x\to7}f(x)=8

indicates that as the input xx approaches 77 from either the left or the right, the output approaches 8.8. The output can get as close to 88 as we like if the input is sufficiently near 7.7.

What happens at x=7?x=7? When x=7,x=7, there is no corresponding output. We write this as

f(7) does not exist.f(7)\text{ does not exist.}

This notation indicates that 77 is not in the domain of the function. We had already indicated this when we wrote the function as

f(x)=x+1, x7.f(x)=x+1,\ x\ne7.

Notice that the limit of a function can exist even when f(x)f(x) is not defined at x=a.x=a. Much of our subsequent work will be determining limits of functions as xx nears a,a, even though the output at x=ax=a does not exist.

The Limit of a Function. A quantity LL is the limit of a function f(x)f(x) as xx approaches aa if, as the input values of xx approach aa (but do not equal aa), the corresponding output values of f(x)f(x) get closer to L.L. Note that the value of the limit is not affected by the output value of f(x)f(x) at a.a. Both aa and LL must be real numbers. We write it as

limxaf(x)=L\lim\limits_{x\to a}f(x)=L

Example. For the following limit, define a,a, f(x),f(x), and L.L.

limx2(3x+5)=11\lim\limits_{x\to2}(3x+5)=11

Solution. First, we recognize the notation of a limit. If the limit exists, as xx approaches a,a, we write

limxaf(x)=L.\lim\limits_{x\to a}f(x)=L.

We are given

limx2(3x+5)=11.\lim\limits_{x\to2}(3x+5)=11.

This means that a=2,a=2, f(x)=3x+5,f(x)=3x+5, and L=11.L=11.

Analysis. Recall that y=3x+5y=3x+5 is a line with no breaks. As the input values approach 2,2, the output values will get close to 11.11. This may be phrased with the equation limx2(3x+5)=11,\lim_{x\to2}(3x+5)=11, which means that as xx nears 22 (but is not exactly 22), the output of the function f(x)=3x+5f(x)=3x+5 gets as close as we want to 3(2)+5,3(2)+5, or 11,11, which is the limit L,L, as we take values of xx sufficiently near 22 but not at x=2.x=2.

For the limitlimx5(2x24)=46\lim_{x\to5}(2x^2-4)=46, what isaa?

For the limitlimx5(2x24)=46\lim_{x\to5}(2x^2-4)=46, what isf(x)f(x)?

For the limitlimx5(2x24)=46\lim_{x\to5}(2x^2-4)=46, what isLL?

Understanding Left-Hand Limits and Right-Hand Limits

We can approach the input of a function from either side of a value—from the left or the right. The table below shows the values of

f(x)=x+1, x7f(x)=x+1,\ x\ne7

as described earlier.

xx6.96.96.996.996.9996.999777.0017.0017.017.017.17.1
f(x)f(x)7.97.97.997.997.9997.999undefined8.0018.0018.018.018.18.1

Values described as “from the left” are less than the input value 77 and would therefore appear to the left of the value on a number line. The input values that approach 77 from the left in the table are 6.9,6.9, 6.99,6.99, and 6.999.6.999. The corresponding outputs are 7.9,7.9, 7.99,7.99, and 7.999.7.999. These values are getting closer to 8.8. The limit of values of f(x)f(x) as xx approaches from the left is known as the left-hand limit. For this function, 88 is the left-hand limit of the function f(x)=x+1, x7f(x)=x+1,\ x\ne7 as xx approaches 7.7.

Values described as “from the right” are greater than the input value 77 and would therefore appear to the right of the value on a number line. The input values that approach 77 from the right in the table are 7.1,7.1, 7.01,7.01, and 7.001.7.001. The corresponding outputs are 8.1,8.1, 8.01,8.01, and 8.001.8.001. These values are getting closer to 8.8. The limit of values of f(x)f(x) as xx approaches from the right is known as the right-hand limit. For this function, 88 is also the right-hand limit of the function f(x)=x+1, x7f(x)=x+1,\ x\ne7 as xx approaches 7.7.

The table shows that we can get the output of the function within a distance of 0.10.1 from 88 by using an input within a distance of 0.10.1 from 7.7. In other words, we need an input xx within the interval 6.9<x<7.16.9<x<7.1 to produce an output value of f(x)f(x) within the interval 7.9<f(x)<8.1.7.9<f(x)<8.1.

We also see that we can get output values of f(x)f(x) successively closer to 88 by selecting input values closer to 7.7. In fact, we can obtain output values within any specified interval if we choose appropriate input values.

The graph below provides a visual representation of the left- and right-hand limits of the function. From the graph of f(x),f(x), we observe the output can get infinitesimally close to L=8L=8 as xx approaches 77 from the left and as xx approaches 77 from the right.

To indicate the left-hand limit, we write

limx7f(x)=8.\lim\limits_{x\to7^-}f(x)=8.

To indicate the right-hand limit, we write

limx7+f(x)=8.\lim\limits_{x\to7^+}f(x)=8.

Left- and Right-Hand Limits. The left-hand limit of a function f(x)f(x) as xx approaches aa from the left is equal to L,L, denoted by

limxaf(x)=L.\lim\limits_{x\to a^-}f(x)=L.

The values of f(x)f(x) can get as close to the limit LL as we like by taking values of xx sufficiently close to aa such that x<ax<a and xa.x\ne a.

The right-hand limit of a function f(x),f(x), as xx approaches aa from the right, is equal to L,L, denoted by

limxa+f(x)=L.\lim\limits_{x\to a^+}f(x)=L.

The values of f(x)f(x) can get as close to the limit LL as we like by taking values of xx sufficiently close to aa but greater than a.a. Both aa and LL are real numbers.

Understanding Two-Sided Limits

In the previous example, the left-hand limit and right-hand limit as xx approaches aa are equal. If the left- and right-hand limits are equal, we say that the function f(x)f(x) has a two-sided limit as xx approaches a.a. More commonly, we simply refer to a two-sided limit as a limit. If the left-hand limit does not equal the right-hand limit, or if one of them does not exist, we say the limit does not exist.

The Two-Sided Limit of a Function as xx Approaches aa. The limit of a function f(x),f(x), as xx approaches a,a, is equal to L,L, that is,

limxaf(x)=L\lim\limits_{x\to a}f(x)=L

if and only if

limxaf(x)=limxa+f(x).\lim\limits_{x\to a^-}f(x)=\lim\limits_{x\to a^+}f(x).

In other words, the left-hand limit of a function f(x)f(x) as xx approaches aa is equal to the right-hand limit of the same function as xx approaches a.a. If such a limit exists, we refer to the limit as a two-sided limit. Otherwise we say the limit does not exist.

Finding a Limit Using a Graph

To visually determine if a limit exists as xx approaches a,a, we observe the graph of the function when xx is very near to x=a.x=a. In the graph below we observe the behavior of the graph on both sides of a.a.

To determine if a left-hand limit exists, we observe the branch of the graph to the left of x=a,x=a, but near x=a.x=a. This is where x<a.x<a. We see that the outputs are getting close to some real number LL so there is a left-hand limit.

To determine if a right-hand limit exists, observe the branch of the graph to the right of x=a,x=a, but near x=a.x=a. This is where x>a.x>a. We see that the outputs are getting close to some real number L,L, so there is a right-hand limit.

If the left-hand limit and the right-hand limit are the same, as they are above, then we know that the function has a two-sided limit. Normally, when we refer to a “limit,” we mean a two-sided limit, unless we call it a one-sided limit.

Finally, we can look for an output value for the function f(x)f(x) when the input value xx is equal to a.a. The coordinate pair of the point would be (a,f(a)).(a,f(a)). If such a point exists, then f(a)f(a) has a value. If the point does not exist, as above, then we say that f(a)f(a) does not exist.

How To: Given a function f(x),f(x), use a graph to find the limits and a function value as xx approaches a.a.

  1. Examine the graph to determine whether a left-hand limit exists.
  2. Examine the graph to determine whether a right-hand limit exists.
  3. If the two one-sided limits exist and are equal, then there is a two-sided limit—what we normally call a “limit.”
  4. If there is a point at x=a,x=a, then f(a)f(a) is the corresponding function value.

Example. Determine the following limits and function value for the function ff shown in the first graph below: limx2f(x),\lim_{x\to2^-}f(x), limx2+f(x),\lim_{x\to2^+}f(x), limx2f(x),\lim_{x\to2}f(x), and f(2).f(2).

Then determine the same four quantities for the function ff shown in the second graph below.

Solution. Looking at the first graph:

  • limx2f(x)=8;\lim_{x\to2^-}f(x)=8; when x<2,x<2, but infinitesimally close to 2,2, the output values get close to y=8.y=8.
  • limx2+f(x)=3;\lim_{x\to2^+}f(x)=3; when x>2,x>2, but infinitesimally close to 2,2, the output values approach y=3.y=3.
  • limx2f(x)\lim_{x\to2}f(x) does not exist because limx2f(x)limx2+f(x);\lim_{x\to2^-}f(x)\ne\lim_{x\to2^+}f(x); the left- and right-hand limits are not equal.
  • f(2)=3f(2)=3 because the graph of the function ff passes through the point (2,f(2))(2,f(2)) or (2,3).(2,3).

Looking at the second graph:

  • limx2f(x)=8;\lim_{x\to2^-}f(x)=8; when x<2x<2 but infinitesimally close to 2,2, the output values approach y=8.y=8.
  • limx2+f(x)=8;\lim_{x\to2^+}f(x)=8; when x>2x>2 but infinitesimally close to 2,2, the output values approach y=8.y=8.
  • limx2f(x)=8\lim_{x\to2}f(x)=8 because limx2f(x)=limx2+f(x)=8;\lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)=8; the left- and right-hand limits are equal.
  • f(2)=4f(2)=4 because the graph of the function ff passes through the point (2,f(2))(2,f(2)) or (2,4).(2,4).

Using the graph of y=f(x)y=f(x) shown below, estimate the following limits: limx0f(x),\lim_{x\to0^-}f(x), limx0+f(x),\lim_{x\to0^+}f(x), limx0f(x),\lim_{x\to0}f(x), limx2f(x),\lim_{x\to2^-}f(x), limx2+f(x),\lim_{x\to2^+}f(x), limx2f(x),\lim_{x\to2}f(x), limx4f(x),\lim_{x\to4^-}f(x), limx4+f(x),\lim_{x\to4^+}f(x), and limx4f(x).\lim_{x\to4}f(x).

Using the graph above, estimatelimx0f(x)\lim_{x\to0^-}f(x).

Using the same graph, determinelimx2f(x)\lim_{x\to2}f(x).

Using the same graph, determinelimx4f(x)\lim_{x\to4}f(x).

Finding a Limit Using a Table

Creating a table is a way to determine limits using numeric information. We create a table of values in which the input values of xx approach aa from both sides. Then we determine if the output values get closer and closer to some real value, the limit L.L.

Let’s consider an example using the following function:

limx5(x3125x5)\lim\limits_{x\to5}\left(\tfrac{x^3-125}{x-5}\right)

To create the table, we evaluate the function at values close to x=5.x=5. We use some input values less than 55 and some values greater than 5,5, as in the table below. The table values show that when x<5x<5 but nearing 5,5, the corresponding output gets close to 75.75. When x>5x>5 but nearing 5,5, the corresponding output also gets close to 75.75.

xx4.94.94.994.994.9994.999555.0015.0015.015.015.15.1
f(x)f(x)73.5173.5174.850174.850174.98500174.985001undefined75.01500175.01500175.150175.150176.5176.51

Because

limx5f(x)=75=limx5+f(x),\lim\limits_{x\to5^-}f(x)=75=\lim\limits_{x\to5^+}f(x),

then

limx5f(x)=75.\lim\limits_{x\to5}f(x)=75.

Remember that f(5)f(5) does not exist.

How To: Given a function f,f, use a table to find the limit as xx approaches aa and the value of f(a),f(a), if it exists.

  1. Choose several input values that approach aa from both the left and right. Record them in a table.
  2. Evaluate the function at each input value. Record them in the table.
  3. Determine if the table values indicate a left-hand limit and a right-hand limit.
  4. If the left-hand and right-hand limits exist and are equal, there is a two-sided limit.
  5. Replace xx with aa to find the value of f(a).f(a).

Example. Numerically estimate the limit of the following expression by setting up a table of values on both sides of the limit.

limx0(5sin(x)3x)\lim\limits_{x\to0}\left(\tfrac{5\sin(x)}{3x}\right)

Solution. We can estimate the value of a limit, if it exists, by evaluating the function at values near x=0.x=0. We cannot find a function value for x=0x=0 directly because the result would have a denominator equal to 0,0, and thus would be undefined.

f(x)=5sin(x)3xf(x)=\tfrac{5\sin(x)}{3x}

We create the table below by choosing several input values close to x=0,x=0, with half of them less than x=0x=0 and half of them greater than x=0.x=0. Note that we need to be sure we are using radian mode. We evaluate the function at each input value to complete the table.

The table values indicate that when x<0x<0 but approaching 0,0, the corresponding output nears 53.\tfrac{5}{3}.

When x>0x>0 but approaching 0,0, the corresponding output also nears 53.\tfrac{5}{3}.

xx0.1-0.10.01-0.010.001-0.001000.0010.0010.010.010.10.1
f(x)f(x)1.663891.663891.6666391.6666391.6666661.666666undefined1.6666661.6666661.6666391.6666391.663891.66389

Because

limx0f(x)=53=limx0+f(x),\lim\limits_{x\to0^-}f(x)=\tfrac{5}{3}=\lim\limits_{x\to0^+}f(x),

then

limx0f(x)=53.\lim\limits_{x\to0}f(x)=\tfrac{5}{3}.

Numerically estimate the limitlimx0(20sin(x)4x)\lim_{x\to0}\left(\tfrac{20\sin(x)}{4x}\right)by making a table of values.

Q&A. Is it possible to check our answer using a graphing utility?

Yes. We previously used a table to find a limit of 7575 for the function f(x)=x3125x5f(x)=\tfrac{x^3-125}{x-5} as xx approaches 5.5. To check, we graph the function on a viewing window as shown below. A graphical check shows both branches of the graph of the function get close to the output 7575 as xx nears 5.5. Furthermore, we can use the “trace” feature of a graphing calculator. By approaching x=5x=5 we may numerically observe the corresponding outputs getting close to 75.75.

Q&A. Is one method for determining a limit better than the other?

No. Both methods have advantages. Graphing allows for quick inspection. Tables can be used when graphical utilities aren’t available, and they can be calculated to a higher precision than could be seen with an unaided eye inspecting a graph.

Example. With the use of a graphing utility, if possible, determine the left- and right-hand limits of the following function as xx approaches 0.0. If the function has a limit as xx approaches 0,0, state it. If not, discuss why there is no limit.

f(x)=3sin(πx)f(x)=3\sin\left(\tfrac{\pi}{x}\right)

Solution. We can use a graphing utility to investigate the behavior of the graph close to x=0.x=0. Centering around x=0,x=0, we choose two viewing windows such that the second one is zoomed in closer to x=0x=0 than the first one. The result would resemble the graph below for [2,2][-2,2] by [3,3].[-3,3].

The result would resemble the graph below for [0.1,0.1][-0.1,0.1] by [3,3].[-3,3].

The closer we get to 0,0, the greater the swings in the output values are. That is not the behavior of a function with either a left-hand limit or a right-hand limit. And if there is no left-hand limit or right-hand limit, there certainly is no limit to the function f(x)f(x) as xx approaches 0.0.

We write

limx0(3sin(πx)) does not exist.\lim\limits_{x\to0^-}\left(3\sin\left(\tfrac{\pi}{x}\right)\right)\text{ does not exist.}limx0+(3sin(πx)) does not exist.\lim\limits_{x\to0^+}\left(3\sin\left(\tfrac{\pi}{x}\right)\right)\text{ does not exist.}limx0(3sin(πx)) does not exist.\lim\limits_{x\to0}\left(3\sin\left(\tfrac{\pi}{x}\right)\right)\text{ does not exist.}

Numerically estimate the limitlimx0sin(2x)\lim_{x\to0}\sin\left(\tfrac{2}{x}\right).

Media. Access these online resources for additional instruction and practice with finding limits.

Key concepts

  • A function has a limit if the output values approach some value LL as the input values approach some quantity a.a.
  • A shorthand notation is used to describe the limit of a function according to the form limxaf(x)=L,\lim_{x\to a}f(x)=L, which indicates that as xx approaches a,a, both from the left of x=ax=a and the right of x=a,x=a, the output value gets close to L.L.
  • A function has a left-hand limit if f(x)f(x) approaches LL as xx approaches aa where x<a.x<a. A function has a right-hand limit if f(x)f(x) approaches LL as xx approaches aa where x>a.x>a.
  • A two-sided limit exists if the left-hand limit and the right-hand limit of a function are the same. A function is said to have a limit if it has a two-sided limit.
  • A graph provides a visual method of determining the limit of a function.
  • If the function has a limit as xx approaches a,a, the branches of the graph will approach the same yy-coordinate near x=ax=a from the left and the right.
  • A table can be used to determine if a function has a limit. The table should show input values that approach aa from both directions so that the resulting output values can be evaluated. If the output values approach some number, the function has a limit.
  • A graphing utility can also be used to find a limit.

Practice

Understand limit notation

Explain the difference between a value atx=ax=aand the limit asxxapproachesaa.

For the limitlimx3(2x2+1)=19\lim_{x\to-3}(2x^2+1)=19, what isaa?

For the limitlimx3(2x2+1)=19\lim_{x\to-3}(2x^2+1)=19, what isf(x)f(x)?

For the limitlimx3(2x2+1)=19\lim_{x\to-3}(2x^2+1)=19, what isLL?

Find a limit using a graph

For the following exercises, estimate the functional values and the limits from the graph of the function ff shown below.

Estimatelimx2f(x)\lim_{x\to-2^-}f(x)from the graph above.

Estimatelimx2f(x)\lim_{x\to-2}f(x)from the graph above.

Estimatelimx1f(x)\lim_{x\to1^-}f(x)from the graph above.

Estimatelimx1f(x)\lim_{x\to1}f(x)from the graph above.

Estimatelimx4f(x)\lim_{x\to4^-}f(x)from the graph above.

Estimatelimx4f(x)\lim_{x\to4}f(x)from the graph above.

Find a limit using a table

Use numerical evidence to estimatelimx3(x2x6x29)\lim_{x\to3}\left(\tfrac{x^2-x-6}{x^2-9}\right). Round your answer to two decimal places.

Use numerical evidence to estimatelimx1(x21x23x+2)\lim_{x\to1}\left(\tfrac{x^2-1}{x^2-3x+2}\right). Round your answer to two decimal places.

Use numerical evidence to estimatelimx1(1010x2x23x+2)\lim_{x\to1}\left(\tfrac{10-10x^2}{x^2-3x+2}\right). Round your answer to two decimal places.

Use numerical evidence to determine whetherlimx12(x4x2+4x+1)\lim_{x\to-\frac12}\left(\tfrac{x}{4x^2+4x+1}\right)exists.

Use a calculator to estimatelimx07tanx3x\lim_{x\to0}\tfrac{7\tan x}{3x}by preparing a table of values, as a fraction.

Use a graphing utility to find numerical evidence to estimatelimx0ee1x2\lim_{x\to0}e^{e^{-\frac{1}{x^2}}}.


This section is adapted from Precalculus 2e, Section 12.1: Finding Limits: Numerical and Graphical Approaches by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: Every source figure is recreated as an apfigure spec, never a traced image. The two “generic” limit-at-(a,L)(a,L) illustrations (the intro figure and the “finding a limit using a graph” figure) have no named source formula, so each was fitted by script to a low-degree polynomial matching the open circle at (a,L)(a,L): the intro S-curve is the cubic y=0.12(x1)3+2y=0.12(x-1)^3+2, and the hump figure is the quadratic y=0.4(x1)2+3y=-0.4(x-1)^2+3. The hole graph of f(x)=x26x7x7=x+1, x7f(x)=\tfrac{x^2-6x-7}{x-7}=x+1,\ x\ne7 (bare, then annotated with the y=8y=8 guide line) is the line of slope 11 with an open circle at (7,8)(7,8). Example 2’s two piecewise graphs are each an upward parabola y=2x2y=2x^2 (drawn from x=4x=-4) meeting a line or a second open circle at x=2x=2, with coordinates read directly off the printed page. The Try It graph (the nine-part graph read after Example 2) is a quadratic y=0.42x2+1.092x+0.01y=0.42x^2+1.092x+0.01 ending open at the origin, a second quadratic y=x2+2y=-x^2+2 running from a filled point at (0,2)(0,2) to a filled point at (2,2)(2,-2), and a straight line of slope 22 from an open circle at (2,0)(2,0) through an open circle at (4,4)(4,4), so every open and filled dot lands exactly on the values the printed answer key gives (0,2,2,0,40,2,-2,0,4); only limx0f(x)\lim_{x\to0^-}f(x), limx2f(x)\lim_{x\to2}f(x), and limx4f(x)\lim_{x\to4}f(x) were carried into real components — the other six of its nine parts (the right-hand limits at 00 and 44, the individual one-sided limits at x=2x=2, and the two two-sided-limit-exists parts already implied by the carried items) were not, per the playbook’s nine-part-Try-It allowance. The graphing-utility check graph in the Q&A is the algebraic simplification of x3125x5\tfrac{x^3-125}{x-5}, namely x2+5x+25x^2+5x+25, with an open circle at (5,75)(5,75); it duplicates Example 3’s function because the source’s own Q&A duplicates it as a graphical check. The two 3sin(π/x)3\sin(\pi/x) zoom windows are not a named curve family, so each was sampled directly from the formula into a pair of polylines (roughly 500 points per branch, excluding a small neighborhood of the x=0x=0 singularity); the resulting aliasing into a dense packed band near the center in both windows reproduces the same compressed look the printed figure shows, rather than smoothing it away. The graphical-exercise figure (discontinuities at x=2,1,4x=-2,1,4) was transcribed dot-for-dot from the printed page: a steep line segment for x<2x<-2, a downward parabola y=2x2+4y=-2x^2+4 from an open circle at (2,4)(-2,-4) to an open circle at (1,2)(1,2) with an isolated filled point at (2,3)(-2,3), a second open circle at (1,0)(1,0) starting a square-root branch y=2.309x1y=2.309\sqrt{x-1} through a filled point at (4,4)(4,4), and a short separate hooked piece starting from an open circle at (4,6)(4,6) that no exercise reads numerically. Corrected one spelling slip in the graphing-utility Q&A (the source prints “appraoching”), a non-mathematical transcription slip. Omitted the Media callout’s two external resource links, keeping its introductory sentence, matching house precedent elsewhere in this book. Every retained Try It is a real fillin or multiplechoice component; every “does not exist” answer is multiplechoice, since MathLive cannot type words and a fillin cannot key a bare-text answer; every “evaluate/estimate the limit” fillin declares decimal or fraction so the engine’s own evaluation of the printed limit expression cannot pass as a retype (verified against the real grader). The “Understand limit notation” Practice group draws its multiplechoice from the section’s Verbal exercises (fs-id1165137810884, the only one of the two with a printed solution) and, since that leaves the group with only one item, adds a fresh “identify aa, f(x)f(x), and LL” triple in the pattern of Example 1 / Try It 1 with a new substitution, limx3(2x2+1)=19\lim_{x\to-3}(2x^2+1)=19, not printed in the source, per the brief’s allowance for a thin objective. The “Find a limit using a graph” group is the six graph-read exercises with printed solutions from the “Graphical” set reading the discontinuities figure (fs-id1165135183014, fs-id1165137806213, fs-id1165137935628, fs-id1165135181688, fs-id1165137455877, fs-id1165137794235 — printed numbers 3, 5, 7, 9, 11, 13). The “Find a limit using a table” group is six exercises with printed solutions from the “Numeric” set (fs-id1165135195366, fs-id1165137836967, fs-id1165137770148, fs-id1165135209740, fs-id1165135188629, fs-id1165137731539 — printed numbers 31, 33, 35, 37, 39, 43); the last of those (printed #43) belongs to the “use a graphing utility to find numerical or graphical evidence” subset rather than the “Numeric” subset proper, but is the same kind of table-driven estimate and is independently re-derived like every other item here. The Extensions subsection’s relativistic-mass conjecture exercise and its accompanying table (Table_12_01_01) were not used: that item asks for an open-ended conjecture rather than a single checkable value, and the two Practice groups above already meet the “find a limit using a graph/table” objectives without it.