Precalculus by Richard Wright

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# 12-01 Introduction to Limits

Summary: In this section, you will:

• Evaluate limits with a table.
• Identify when limits fail to exist.
• Evaluate limits by substitution.

SDA NAD Content Standards (2018): PC.4.2

The speed limit sign like the one above are common to see along roads. The sign gives the maximum speed that you can legally drive on that road. You can legally approach that speed, but you cannot go over that speed. Mathematical limits are a similar concept.

## Limits

In calculus, a limit is the value that a function approaches as you get arbitrarily near a x-value. Unlike a speed limit, you do not necessarily reach the value of the limit, but you get close enough to it to know what that value will be. It would be like driving your car 54.999999 mph. You are not at the limit of 55 mph, but you are close enough.

The formal definition of a limit is if f(x) becomes arbitrarily close to a unique number L as x approaches c from either side, then the limit of f(x) as x approaches c is L. It is written as

$$\lim_{x \rightarrow c} f(x) = L$$

This is read "The limit as x approaches c of f(x) is L." It means that as x gets very close to c from both sides, then f(x) gets very close to L. Thus, the limit is L.

###### Limit

if f(x) becomes arbitrarily close to a unique number L as x approaches c from either side, then the limit of f(x) as x approaches c is L.

$$\lim_{x \rightarrow c} f(x) = L$$

One way to calculate a limit is to make a table. Unlike the usual table that is often used in math, the x values are going to get closer to c by decimal places. For example, if you want the limit at 2, use the x-values such as 1.9, 1.99, 1.999, 1.9999, 2.0001, 2.001, 2.01, 2.1. This will let you see a pattern as you approach 2.

#### Example 1: Evaluate a Limit Using a Table

Use a table to evaluate $$\displaystyle \lim_{x \rightarrow 3} \frac{x^2 - 4x + 3}{x - 3}$$

###### Solution

The limit is near x = 3, so make a table near 3.

 x f(x) 2.9 2.99 2.999 2.9999 3 3.0001 3.001 3.01 3.1 1.9 1.99 1.999 1.9999 __?__ 2.0001 2.001 2.01 2.1

It appears from the pattern that the limit approaches 2. This can also be seen in the graph below. If there was not a hole in the graph, it would go through the point (3, 2).

$$\lim_{x \rightarrow 3} \frac{x^2 - 4x + 3}{x - 3} = 2$$

##### Try It 1

Use a table to evaluate $$\displaystyle \lim_{x \rightarrow 0} \frac{2x^3 - 3x}{x}$$

–3

## Limits that Fail to Exist

The graph of the function does not necessarily need to go through a point for the limit to exist. The function in Example 1 has a hole in it and is undefined when x = 3, but the graph does approach 2 as x approaches 3. However, there are some limits that do not exist. There are three cases: (1) the function approaches different numbers from either side, (2) the function increases or decreases without bound, or (3) the function oscillates between two fixed values.

###### Limits that Do Not Exist
1. f(x) approaches different numbers from either side
2. f(x) increases or decreases without bound
3. f(x) oscillates between 2 fixed values like $$f(x) = \cos \left(\frac{1}{x}\right)$$

#### Example 2: Limits that Do Not Exist

State whether the limit exists. If it does not, explain why not.

1. $$\displaystyle \lim_{x \rightarrow 1} \frac{2x^2 + 3x + 1}{x - 1}$$
2. $$\displaystyle \lim_{x \rightarrow 0} \sin \left(\frac{1}{x}\right)$$
3. $$\displaystyle \lim_{x \rightarrow -2} | x + 2 |$$
4. $$\displaystyle \lim_{x \rightarrow -3} \frac{| x + 3 |}{x + 3}$$
###### Solution

Make a quick graph on your calculator and look at the value x approaches.

1. At x = 1, the graph goes both down towards –∞ from the left and up towards +∞ from the right. The limit does not exist because the function increases or decreases without bound.

2. At x = 0, the graph oscillates between –1 and 1. Zoom in several times to see what happens. The limit does not exist because the function oscillates between two values.

3. At x = –2, the graph approaches 0 from both sides, so the limit is 0.

4. At x = –3, the graph approaches –1 from the left and 1 from the right. The limit does not exist because the function approaches different values from either side.

##### Try It 2

State whether the limit exists. If it does not, explain why not. $$\displaystyle \lim_{x \rightarrow 0} \frac{2}{x^2}$$

Does not exist because it increases without bound at x = 0.

## Properties of Limits

Many limits can be evaluated simply by plugging in the value x approaches into the function. This is called direct substitution. The first few properties in the following list illustrate this. The other properties show how to combine limits.

###### Properties of Limits
• $$\displaystyle \lim_{x \rightarrow c} b = b$$
• $$\displaystyle \lim_{x \rightarrow c} x = c$$
• $$\displaystyle \lim_{x \rightarrow c} x^n = c^n$$
• $$\displaystyle \lim_{x \rightarrow c} \sqrt[n]{x} = \sqrt[n]{c}$$

Let $$\displaystyle \lim_{x \rightarrow c} f(x) = L$$ and $$\displaystyle \lim_{x \rightarrow c} g(x) = K$$.

• $$\displaystyle \lim_{x \rightarrow c} bf(x) = bL$$
• $$\displaystyle \lim_{x \rightarrow c} [f(x) ± g(x)] = L ± K$$
• $$\displaystyle \lim_{x \rightarrow c} f(x)g(x) = LK$$
• $$\displaystyle \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \frac{L}{K}$$
• $$\displaystyle \lim_{x \rightarrow c} [f(x)]^n = L^n$$

#### Example 3: Evaluate a Limit

Evaluate $$\displaystyle \lim_{x \rightarrow 2} 4x^3$$

###### Solution

The properties can be summarized as try to plug in the value x approaches. So in this case, plug in a 2.

$$\lim_{x \rightarrow 2} 4x^3$$

$$= 4(2)^3$$

$$= 32$$

#### Example 4: Evaluate a Limit

Evaluate $$\displaystyle \lim_{x \rightarrow -1} (2x^2 - 3x + 5)$$.

###### Solution

x approaches −1, so try plugging it in.

$$\lim_{x \rightarrow -1} (2x^2 - 3x + 5)$$

$$= 2(-1)^2 - 3(-1) + 5$$

$$= 10$$

#### Example 5: Evaluate a Limit

Evaluate $$\displaystyle \lim_{x \rightarrow 0} \frac{\sqrt{x + 1}}{2x}$$.

###### Solution

x approaches 0, so try plugging in a 0.

$$\lim_{x \rightarrow 0} \frac{\sqrt{x + 1}}{2x}$$

$$= \frac{\sqrt{0 + 1}}{2(0)}$$

$$= \frac{1}{0}$$

Since you cannot divide by 0, this limit Does not exist.

##### Try It 3

Evaluate $$\displaystyle \lim_{x \rightarrow 1} \frac{x^2 - 2x}{x}$$

–1 (This exists because the function is defined at x = 1. $$\displaystyle \lim_{x \rightarrow 0} \frac{x^2 - 2x}{x}$$ does not exist because it increases without bound at x = 0.)

##### Lesson Summary

###### Limit

if f(x) becomes arbitrarily close to a unique number L as x approaches c from either side, then the limit of f(x) as x approaches c is L.

$$\lim_{x \rightarrow c} f(x) = L$$

###### Limits that Do Not Exist
1. f(x) approaches different numbers from either side
2. f(x) increases or decreases without bound
3. f(x) oscillates between 2 fixed values like $$f(x) = \cos \left(\frac{1}{x}\right)$$

###### Properties of Limits
• $$\displaystyle \lim_{x \rightarrow c} b = b$$
• $$\displaystyle \lim_{x \rightarrow c} x = c$$
• $$\displaystyle \lim_{x \rightarrow c} x^n = c^n$$
• $$\displaystyle \lim_{x \rightarrow c} \sqrt[n]{x} = \sqrt[n]{c}$$

Let $$\displaystyle \lim_{x \rightarrow c} f(x) = L$$ and $$\displaystyle \lim_{x \rightarrow c} g(x) = K$$.

• $$\displaystyle \lim_{x \rightarrow c} bf(x) = bL$$
• $$\displaystyle \lim_{x \rightarrow c} [f(x) ± g(x)] = L ± K$$
• $$\displaystyle \lim_{x \rightarrow c} f(x)g(x) = LK$$
• $$\displaystyle \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \frac{L}{K}$$
• $$\displaystyle \lim_{x \rightarrow c} [f(x)]^n = L^n$$

## Practice Exercises

Use a table to evaluate the limit.

1. $$\displaystyle \lim_{x \rightarrow -4} \frac{x^2 + 3x - 4}{x + 4}$$
2. $$\displaystyle \lim_{x \rightarrow 0} \frac{3x^2 + 4x}{2x}$$
3. \displaystyle \lim_{x \rightarrow -1} \left\{\begin{align} 2x + 1, \quad &x ≤ -1 \\ x^2 - 2, \quad &x > -1 \end{align} \right.
4. Use a graph to determine if the limit exists. If it does not, explain why. If it does, evaluate the limit.

5. $$\displaystyle \lim_{x \rightarrow 0} 3 \cos \left(\frac{1}{2x} \right)$$
6. $$\displaystyle \lim_{x \rightarrow 0} -\frac{x^3 + 3x}{x^3}$$
7. $$\displaystyle \lim_{x \rightarrow 1} \frac{2x + 1}{3x - 4}$$
8. $$\displaystyle \lim_{x \rightarrow 2} \frac{2\sqrt{x + 2}}{x - 2}$$
9. \displaystyle \lim_{x \rightarrow 0} \left\{ \begin{align} x - 3, \quad &x ≤ 0 \\ |x + 3|, \quad &x > 0 \end{align}\right.
10. Evaluate the limit by direct substitution. If it does not exist, say so.

11. $$\displaystyle \lim_{x \rightarrow 1} \sin\left(\frac{π}{x}\right)$$
12. $$\displaystyle \lim_{x \rightarrow 2} 2x^3 - 5x$$
13. $$\displaystyle \lim_{x \rightarrow 0} 3\sqrt{x + 1}$$
14. $$\displaystyle \lim_{x \rightarrow 2} \frac{x^2 + 3x + 1}{x + 1}$$
15. $$\displaystyle \lim_{x \rightarrow 0} 2x(x - 1)$$
16. \displaystyle \lim_{x \rightarrow 2} \left\{\begin{align} -x^2, \quad &x ≤ 0 \\ x^2, \quad &x > 0\end{align}\right.
17. Problem Solving

18. The population of some sparrows in the backyard can modeled by $$P(t) = \frac{75}{1 + e^{-t + 2}}$$ where P is the number of sparrows in the backyard after t years. Find the limit of the population as time approaches 4 years.
19. Mixed Review

20. (11-04) Find the distance from the point (3, 2, –3) to the plane x + 2y + z – 5 = 0.
21. (11-04) Find the parametric equations of the line passing through (2, 0, –1) and (4, 2, 2).
22. (11-03) Evaluate ⟨2,2,−1⟩ × ⟨−3,0,1⟩.
23. (10-02) Evaluate the summation using the shortcut formulas. $$\displaystyle \sum_{i=1}^{15} i^2 - 3i$$
24. (1-04) Simplify $$\frac{f(x + h) - f(x)}{h}$$ when $$f(x) = 2x^2 - x$$.

1. –5
2. 2
3. –1
4. DNE, oscillates
5. DNE, decreases without bound
6. –3
7. DNE, increases without bound
8. DNE, approaches different values from either side
9. 0
10. 6
11. 3
12. $$\frac{11}{3}$$
13. 0
14. 4
15. 66 sparrows
16. $$\frac{\sqrt{6}}{6}$$
17. \left\{\begin{align} x &= 2t + 2 \\ y &= 2t \\ z &= 3t - 1 \end{align}\right.
18. ⟨2,1,6⟩
19. 880
20. 4x + 2h – 1