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# 8.1: Limits (An Intuitive Approach)

Difficulty Level: At Grade Created by: CK-12

## Learning objectives

• Demonstrate an understanding of the relationship between the tangent line and the limit.
• Demonstrate an understanding of the area between a curve and the x-axis.
• Understand the notion of the limit of a function as an independent variable approaches a finite or infinite value.

## Introduction

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 1:

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 2: 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.

## The One-Sided Limits

The limit in the definition above and from Examples 1 and 2,

limxxof(x)=L\begin{align*}\lim_{x \rightarrow x_o} f(x) = L\end{align*}

is called a two-sided limit because it requires f(x) to get closer and closer to L from both sides of x = x0.However, not all functions behave this way. Some actually have different limits on the two sides of x = x0. For example, consider the function

f(x)=|x|x={1,x>01,x<0\begin{align*}f(x) = \frac{|x|} {x} = \begin{cases} 1, x > 0\\ -1, x < 0\\ \end{cases} \end{align*}

which is shown in the graph below:

Note that as x approaches 0 from the right, f(x) approaches 1. On the other hand, as x approaches 0 from the left, the function f(x) approaches -1. Mathematically, we write

limx0+|x|x=1\begin{align*}\lim_{x \rightarrow 0^+} \frac{|x|} {x} = 1\end{align*}
limx0|x|x=1\begin{align*}\lim_{x \rightarrow 0^-} \frac{|x|} {x} = -1\end{align*}

Where the superscript “+” indicates a limit from the right and the superscript “-” indicates a limit from the left.

The One-Sided Limit

If f(x) approaches L as x approaches x0from the left and from the right, then we write

limxx+0f(x)=L\begin{align*}\lim_{x \rightarrow x_0^+} f(x) = L\end{align*}

limxx0f(x)=L\begin{align*}\lim_{x \rightarrow x_0^-} f(x) = L\end{align*}

which reads: “the limit of f(x) as x approaches x+0\begin{align*}x_0^+\end{align*} (or x0\begin{align*}x_0^-\end{align*}) from the right (or left) is L.

Sometimes the value of f(x) does not get closer and closer to some single value L as xx0\begin{align*}x \rightarrow x_0\end{align*}. In this case we say that the limit as x approaches x0 does not exist. For example, the two-sided limit limx0|x|x\begin{align*}lim_{x \rightarrow 0} \frac{ |x|}{x}\end{align*} that we have just encountered does not exist because the values of f(x) do not approach a single number as x approaches 0. Actually, the values approach -1 from the left and 1 from the right. This leads to an important condition that must be satisfied in order for the limit of a function to exist. It is shown in the box below.

Conditions For a Limit to Exist (The relationship between one-sided and two-sided limits).

In order for the limit L of a function to exist, both of the one-sided limits must exist at x0 and must have the samevalue. Mathematically,

limxx0f(x)=L\begin{align*}\lim_{x \rightarrow x_0} f(x) = L\end{align*} if and only if limxx0f(x)=L\begin{align*}\lim_{x \rightarrow x_0^-} f(x) = L\end{align*} and limxx0f(x)=L\begin{align*}\lim_{x \rightarrow x_0^-} f(x) = L\end{align*}.

Example 3:

Consider the function f graphed in the accompanying figure and find

1. limx2f(x)\begin{align*} \lim_{x \rightarrow 2^-}f(x)\end{align*}
2. limx2+f(x)\begin{align*} \lim_{x \rightarrow 2^+}f(x)\end{align*}
3. limx2f(x)\begin{align*}\lim_{x \rightarrow 2}f(x)\end{align*}
4. f(2)\begin{align*}f(2)\end{align*}

Solution:

1. From graph, we can see that, limx2f(x)=2\begin{align*}\lim_{x \rightarrow 2^-}f(x) = -2\end{align*}.
2. We can also see from the graph that limx2+f(x)=4\begin{align*}\lim_{x \rightarrow 2^+}f(x) = 4\end{align*}.
3. Since the limits from the right and the left are not equal (they do not approach a single value L), the limit does not exist. That is, limx2f(x)\begin{align*} \lim_{x \rightarrow 2}f(x) \end{align*} does not exist.
4. f(2)=1\begin{align*}f(2)=1\end{align*}.

## Infinite Limits

As previously discussed, the limit of a function may not exist. For example, for the function f(x) = 1/x (shown in the figures below), as x values are taken closer and closer to 0 from the right, the function increases indefinitely. Also, as x values are taken closer and closer to 0 from the left, the function decreases indefinitely.

We describe these limiting behaviors by writing

limx0+1x=+\begin{align*}\lim_{x \rightarrow 0^+} \frac{1} {x} = + \infty\end{align*}
limx01x=\begin{align*}\lim_{x \rightarrow 0^-} \frac{1} {x} = - \infty\end{align*}

Sometimes we want to know the behavior of f(x) as x increases or decreases without bound. In this case we are interested in the end behavior of the function, a concept that was first discussed in Chapter 1. For example, what is the value of f(x) = 1/x as x increases or decreases without bound? That is,

limx+1x=?\begin{align*}\lim_{x \to +\infty} \frac{1} {x} = ?\end{align*}
limx1x=?\begin{align*}\lim_{x \to -\infty} \frac{1} {x} = ?\end{align*}

As you can see from the graphs (shown below), as x decreases without bound, the values of f(x) = 1/x are negative and get closer and closer to 0. On the other hand, as x increases without bound, the values of f(x) = 1/x are positive and still get closer and closer to 0.

That is,

limx+1x=0\begin{align*}\lim_{x \to +\infty} \frac{1} {x} = 0\end{align*}
limx1x=0\begin{align*}\lim_{x \to -\infty} \frac{1} {x} = 0\end{align*}

Example 4: Use the graph from Example 3 (shown again below) and find the limits

1. limx+f(x)\begin{align*} \lim_{x \to +\infty} f(x)\end{align*}
2. limxf(x)\begin{align*} \lim_{x \to -\infty} f(x)\end{align*}

Solution: As you can see from the graph,

1. limx+f(x)=4\begin{align*} \lim_{x \rightarrow + \infty} f(x) = 4\end{align*}, in other words, as x becomes very large, f(x) stays at the same value, 4.
2. \begin{align*} \lim_{x \rightarrow + \infty} f(x) = -2\end{align*}, which says that as x becomes smaller and smaller, f(x) stays at the same value, -2.

## Lesson Summary

In this lesson we have taken a first look at the fundamental problems of calculus, the tangent line problem and the area problem, by investigating the concept of a limit. While in chapter 1 we focused on limits of functions in the context of their end behavior, in this lesson we started to find the limit of a function as it approaches any point by using tables and graphs. We learned about one-sided limits and discovered that for the limit of a function to exist as it approaches a given point, both one-sided limits of the function must exist as it approaches the same point. In the next lesson we will continue our study of limits and learn more techniques for computing limits.

## Points to Consider

1. What is the benefit of knowing a function's one-sided limits as it approaches a given point?
2. What are examples of functions that have limits that do not exist? Do you see connections between these functions?

## Review Questions

1. Make a conjecture about the value of the limit \begin{align*}\lim_{x \rightarrow 0} \frac{1-cos x} {x^2}\end{align*} where x is in radians. (Hint: Make a table of x - values approaching x from the right and the left.)
2. Make a conjecture about the value of the limit \begin{align*}\lim_{x\rightarrow 1} \frac{ln x} {2x - 2}\end{align*}. (Hint: Make a table of x-values approaching x from the right and the left.)
3. Make a conjecture about the value of the limit \begin{align*}\lim_{x\rightarrow 0} \frac{tan2x} {x}\end{align*} where x is in radians. (Hint: Make a table of x - values approaching x from the right and the left.)
4. Make a graph to evaluate the limit \begin{align*}\lim_{x\rightarrow \infty} \frac{x + 2} {x + 3}\end{align*}.
5. From Problem 4, evaluate \begin{align*}\lim_{x\rightarrow 0} \frac{x + 2} {x + 3} \end{align*}.
6. Make a graph to evaluate the limit \begin{align*}\lim_{x\rightarrow \infty} \frac {1} {\sqrt x} \end{align*}.
7. From Problem 6, evaluate \begin{align*}\lim_{x\rightarrow 0^{+}} \frac {1} {\sqrt x} \end{align*}
8. Let
9. \begin{align*}f(x) = \begin{cases}x-1, & x\leq 3 \\ 3x - 7, & x > 3 \end{cases}\end{align*}

10. and find a.
11. \begin{align*}\lim_{x\rightarrow 3^-} f(x)\end{align*}
12. b.
13. \begin{align*}\lim_{x\rightarrow 3^+} f(x)\end{align*}
14. c.
15. \begin{align*}\lim_{x\rightarrow 3} f(x)\end{align*}
16. Does \begin{align*}\lim_{x\rightarrow 0} sin \left (\frac {1} {x} \right)\end{align*} exist? Explain why or why not.
17. Use a calculator (or, better, a spread sheet) to find the exact value of \begin{align*}\lim_{x\rightarrow 0} (1 + x)^{1/x}\end{align*}

1. 1/2
2. 1/2
3. 2
4. 1
5. 2/3
6. 0
7. \begin{align*}+\infty\end{align*}
8. a. 2 b. 2 c. 2
9. Hint: It is best to graph it first. You will notice that the graph oscillates between two numbers (what are they?) as it approaches zero.
10. e

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Date Created:
Feb 23, 2012