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# 8.1: Definition of a Limit

Difficulty Level: At Grade Created by: CK-12
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Jim was watching his girlfriend run in a track meet. She was in the lead, and starting to pull away from the rest of the pack. Jim recognized a great photo opportunity, and snapped a great shot of Becca just as she rounded the corner and entered the homestretch.

Later, they discussed the race over a victory ice cream, as they admired the photo. "You were really moving, Becca," Jim noted.

"I felt like I was flying!" Becca replied.

"I wonder how fast you were running at the exact time I took the photo?" Jim mused.

"That's easy!" Becca said. "Just take the distance of the race, and divide it by the time it took me to run. Here, hand me your phone, I'll run it through your calculator app, what was my time?"

"Hold on, Becca," Jim interjected. "I don't think that will work. You weren't running the same speed the entire race, so dividing your total distance by your total time isn't much more than an educated guess of your speed the instant I took the pic. Maybe we could use the official race recording, it will be time-stamped, and we could reference the track distance markers..."

"Oh, come on Jim! There is no way we can know what my speed was at that instant!" Becca countered. "Any calculation we come up with is going to be no more than an approximation!"

Is Becca right?

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### Guidance

The discovery of calculus was motivated by two fundamental geometric problems: finding the tangent line to a curve and finding the area of a planar region. In this section, we will show that these two problems are related to a deeper concept of calculus known as the limit of a function.

The Two Fundamental Problems of Calculus that Lead to its Discovery:
1. The Tangent Line Problem: What is the equation of the tangent line to the graph f(x) at point P(x0, y0)?
2. The Area Problem: What is the area under the graph f(x) and the x-axis in the interval [a, b]?

The portion of calculus that deals with the tangent problem is called differential calculus and the portion that deals with the area problem is called integral calculus. In order to solve those two problems, we need to have a more precise understanding of what a tangent line is and what is meant by the area under a curve. Both of these issues require us to understand a deeper concept, the limit of a function.

Tangent Lines and Limits

From your studies in geometry, you know that the tangent line is a line that intersects the circle at one point. However, this definition is not precise when we try to apply it to other kinds of curves. For example, as Figure 1 shows, one can draw a tangent line to a curve yet it cuts the curve at more than one point.

So we need to renew our concept of the tangent line and extend it to apply to curves other than circles. To do so, consider point P on the curve in the figure below. If point Q is any other point on the curve that is different from P, the line that passes through P and Q is called the secant line. Imagine if we move point Q along the curve toward point P, the secant line in this case will “rotate” toward a limiting position at point P. Eventually, the secant line will become a tangent line at point P, as the figure below shows. This is a new concept of the tangent line, where the general notion of a tangent line leads to the concept of limit. We will deal with the tangent line in more detail in lesson 8.3.

Area as a limit

Suppose we are interested in finding the area under the curve of a function on the interval [a, b]. For example, consider function f(x) = (x - 2)3 + 1 (Figure a). Let’s say we want to approximate the area under the curve from x = 1 to x = 3. One way to do it is to inscribe rectangles of equal widths on the interval [1, 3] under the curve and then add the areas of these rectangles (Figure b). Intuition tells us that if we repeat the process using more and more rectangles to fill the gaps under the curve, our approximation will approach the exact value of the area under the curve. So, the limiting value of this approximation is the exact value of the area under the curve. If we denote the width of each rectangle by ∆ x and the value of the area under the curve by A, then as ∆ x approaches zero (the widths of the rectangles get thinner and thinner, and thus less and less gaps), then the area A under the curve will reach an exact value.

What we have seen so far is that the concepts of tangent line and area rest on the notion of limit. In the next sections, we will explore those concepts in more details and show how the limit can help us calculate the rate of change of a given quantity. First, however, we introduce some useful notations.

Definition of a Limit (an informal view)
The notation
limxx0f(x)=L\begin{align*}\lim_{x \rightarrow x_0} f(x) = L\end{align*}
means that as x approaches (or gets very close to) x0, the limit of the function f(x) gets very close to the value L. We first used this notation in Chapter 1.

#### Example A

Make a conjecture about the value of the limit of limx03xx+11\begin{align*}\lim_{x \rightarrow 0} \frac{3x} {\sqrt{x + 1} - 1}\end{align*}.

Solution

Notice that the function f(x)=3xx+11\begin{align*}f(x) = \frac{3x} {\sqrt{x + 1} -1}\end{align*} is not defined at x = 0. The table below shows samples of x-values approaching 0 from the left side and from the right side. In both cases, the values of f(x), calculated to at least 5 decimal places, get closer and closer to 6. Thus our conjecture is that limx03xx+11=6\begin{align*}\lim_{x \rightarrow 0} \frac{3x} {\sqrt{x + 1} - 1} = 6\end{align*}.

x
0 0.00001 0.0001 0.001 0.01
f(x) 5.984962 5.9985 5.99985 5.999985 Undefined 6.000015 6.00015 6.0015 6.014963

Another way of seeing this is to graph f(x) (shown below). Notice that the x-values approach 0 from the left side and from the right side. In both cases, the values of f(x) appear to get closer and closer to 6.

Hence, again our conjecture is that limx03xx+11=6\begin{align*}\lim_{x \rightarrow 0} \frac{3x} {\sqrt{x + 1} - 1} = 6\end{align*}.

#### Example B

Make a conjecture about the value of the limit limx0sinxx\begin{align*}\lim_{x \rightarrow 0} \frac{sin x} {x}\end{align*} where x is in radians.

Solution

The function here is not defined at x = 0. With the help of a computing utility, we can obtain the table below.

x
0 0.01 0.1 0.2
f(x) 0.993347 0.998334 0.999983 Undefined 0.999983 0.998334 0.993347

The data in the table suggest thats limx0sinxx=1\begin{align*}\lim_{x \rightarrow 0} \frac{sin x} {x} = 1\end{align*}. The graph below supports this hypothesis.

#### Example C

Make a conjecture about the value of the limit limx01cosxx2\begin{align*}\lim_{x \rightarrow 0} \frac{1-cos x} {x^2}\end{align*} where x is in radians.

Solution

Enter the expression into your graphing calculator, or use this excellent free one here: https://www.desmos.com/calculator

You should get an image like the one below:

It is clear from the graph that the limit is 1/2.

Concept question wrap-up Technically, Becca is correct. However, using calculus to find the limit of her average speed at shorter and shorter intervals around the time the pic was taken could give Jim an answer that would be very, very close, as accurate as the race timer itself anyway.

### Vocabulary

A tangent line is a line that "just touches" a curve at a given point, and no others.

A limit is a value that represents the edge of what a formula can calculate. Often the limit value itself cannot be calculated directly, but must be inferred by ever-closer values above and below it.

A secant line passes through the tangent point on a curve, and also another point on the same curve.

### Guided Practice

Questions

1) Use a grapher to make a conjecture about the value of the limit limx1lnx2x2\begin{align*}\lim_{x\rightarrow 1} \frac{ln x} {2x - 2}\end{align*}.

2) Use a grapher to make a conjecture about the value of the limit limx0tan2xx\begin{align*}\lim_{x\rightarrow 0} \frac{tan2x} {x}\end{align*}.

3) Use limit notation to write "The limit of f(x) equals the cosine of x, as x approaches 2 from the right."

Solutions

1) Using a graphing calc (or https://www.desmos.com/calculator), the limit of 1/2 is easily located:

2) Using a graphing tool:

The limit is 2

3) limx2+cosx\begin{align*}\lim_{x\to 2^+} cos x\end{align*}

### Practice

Write using limit notation:

1. Write the limit of 4x3+3x24x1\begin{align*}4x^3 + 3x^2 - 4x - 1\end{align*} as x\begin{align*}x \end{align*} approaches a\begin{align*}a\end{align*} from the left.
2. Write the limit of g(z)\begin{align*}g(z)\end{align*} as z\begin{align*} z \end{align*} approaches a\begin{align*} a \end{align*} from the left.
3. Write the limit of g(y)\begin{align*}g(y)\end{align*} as y\begin{align*} y \end{align*} approaches b\begin{align*} b \end{align*} from the left.
4. Write the limit of h(z)\begin{align*}h(z)\end{align*} as z\begin{align*} z \end{align*} approaches 1\begin{align*} -1 \end{align*} from the right.
5. Write the limit of h(y)\begin{align*}h(y)\end{align*} as y\begin{align*} y \end{align*} approaches a\begin{align*} a \end{align*} from the left.
6. Write the limit of h(z)\begin{align*}h(z)\end{align*} as z\begin{align*} z \end{align*} approaches a\begin{align*} a \end{align*}

Solve using a calculator to estimate the limit:

1. limx04x+224x\begin{align*}\lim_{x\to0}\frac{\sqrt{-4x+2} - \sqrt{2}}{4x}\end{align*}
2. limx18x214x62x2\begin{align*}\lim_{x \to-1}\frac{-8x^2 - 14x - 6}{-2x - 2}\end{align*}
3. limx0sec(cosx)\begin{align*}\lim_{x\to0}sec(cos x)\end{align*}
4. limx16522x+251110x32\begin{align*}\lim_{x\to\frac{-16}{5}}\frac{\frac{-2}{2x + 2}- \frac{5}{11}}{-10x - 32}\end{align*}
5. limx52x+552x+5\begin{align*}\lim_{x \to\frac{5}{2}} \frac{\sqrt{x + 5} - \sqrt{5}}{-2x + 5}\end{align*}
6. limx4x2+6x+8x+4\begin{align*}\lim_{x\to -4}\frac{x^2 + 6x + 8}{x + 4}\end{align*}
7. limx13252x+3122x13\begin{align*}\lim_{x \to\frac{-13}{2}}\frac{\frac{-5}{2x + 3} - \frac{1}{2}}{-2x - 13}\end{align*}
8. limx0cot(sinx)\begin{align*}\lim_{x\to0} cot(sin x)\end{align*}
9. limx0x+335x\begin{align*}\lim_{x\to0}\frac{\sqrt{x + 3} - \sqrt3}{-5x}\end{align*}
10. limx0tan(cosx)\begin{align*}\lim_{x\to0} tan(cos x)\end{align*}

Write a formal definition for the following problems:

1. limy2tan(y)=L\begin{align*}\lim_{y\to2} tan (y) = L\end{align*}
2. limx1f(x)=N\begin{align*}\lim_{x\to1} f(x) = N\end{align*}
3. limy1x3+2x2+2x+4=L\begin{align*}\lim_{y\to-1} -x^3 + 2x^2 + 2x + 4 = L\end{align*}

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### Vocabulary Language: English Spanish

TermDefinition
secant A line that intersects a circle in two points.
tangent A line that intersects a circle in exactly one point.
End behavior End behavior is a description of the trend of a function as input values become very large or very small, represented as the 'ends' of a graphed function.
Horizontal Asymptote A horizontal asymptote is a horizontal line that indicates where a function flattens out as the independent variable gets very large or very small. A function may touch or pass through a horizontal asymptote.
limit A limit is the value that the output of a function approaches as the input of the function approaches a given value.
limit notation Limit notation is a way of expressing the fact that a function gets arbitrarily close to a value.
secant line A secant line is a line that joins two points on a curve.
Tangent line A tangent line is a line that "just touches" a curve at a single point and no others.

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