MATH 136 Introduction to Limits Given a function y = f (x) , we wish to describe the behavior of the function as the variable

€

x approaches a particular value

€

a . We should be as specific as possible in describing the behavior of both

€

x and the function values f (x) . For instance,

€

x may be increasing to

€

a (i.e., approaching

€

a from the left

€

x→ a−) or

€

x may be decreasing to

€

a (i.e., approaching

€

a from the right

€

x→ a+). Likewise, the function values may be either increasing or decreasing. Example 1. Evaluate and explain the following limits:

(a)

€

lim

x→ 7π6

−4sin x (b)

€

lim

x→ 7π6

+4sin x (c)

€

limx→ 7π

6

4sin x

Solution. We can evaluate the limits by simply computing

€

4sin(7π /6) and then describe the behavior by observing the graph.

€

lim

x→ 7π6

−4sin x = 4sin(7π /6) = 4 × (−1 / 2) = −2

As

€

x increases to

€

7π /6, then

€

4sin x decreases to –2.

Likewise,

€

lim

x→ 7π6

+4sin x = −2 . But now, as

€

x decreases to

€

7π /6, then

€

4sin x increases to –2.

Because these one-sided limits are equal to the same value and are finite, we say that

€

limx→ 7π

6

4sin x exists and

€

limx→ 7π

6

4sin x = –2 also.

For this last limit we say, as

€

x approaches

€

7π /6, then

€

4sin x approaches –2. If the graph is continuous (i.e., no holes, jumps, asymptotes, etc.), then the y–values will approach the function value f (a) as

€

x approaches

€

a (i.e., limx→ a

f ( x) = f (a) ). We

just need to specify how the y–values increase/decrease to f (a) as

€

x is increasing to

€

a (from the left) and as

€

x is decreasing to

€

a (from the right). It is often helpful to graph the function on your calculator and then use one of several built-in commands to compute the function value. Example 2. Let f (x) = 8x − x2 − 4 . Evaluate and describe the limits:

(a)

€

limx→ 2−

f (x) (b)

€

limx→ 2+

f (x) (c)

€

limx→ 2

f (x)

Solution. Because f (x) = 8x − x2 − 4 is a continuous polynomial, we can evaluate each limit as

€

x approaches 2 by simply evaluating

€

f (2) . Then we can observe the graph to describe the behavior.

After graphing with an appropriate WINDOW, it is easy to observe the behavior of the y–values. Again because f is continuous, the y–values will approach the function value f (2) as

€

x approaches 2. We now explain two ways to compute this function value on your calculator: On TI-84: After graphing the function in Y1, press 2nd TRACE (i.e. CALC). Press 1 for value. Enter 2 for X. Then f (2) is given as Y = 8. On TI-89: After graphing the function in y1 (in APPS, Y= Editor), press F5 for MATH, then press 1 for Value. Enter 2 for xc. Then f (2) is given as yc = 8. Now:

€

limx→ 2−

f (x) = 8 As

€

x increases to 2, then

€

8x − x2 − 4 increases to 8.

€

limx→ 2+

f (x) = 8 As

€

x decreases to 2, then

€

8x − x2 − 4 decreases to 8.

€

limx→ 2

f (x) = 8 As

€

x approaches to 2, then

€

8x − x2 − 4 approaches 8.

Second Method of Function Evaluation

On TI-84: After entering the function in Y1, press 2nd Quit to return to the Home screen. Then press VARS. Scroll right to Y-VARS. Press 1 for Function. Press 1 for Y1. Then Y1 comes up on the Home screen. Finish typing Y1(2) and press ENTER. Then f (2) is computed as 8.

Enter function Return to Home Press VARS Scroll right, press 1

Press 1 for Y1 Enter Y1(2)

On TI-89: After entering the function in y1, enter y1(2) to obtain f (2)= 8.

One-Sided Limits

Formally, we are evaluating limits. To express that

€

x is increasing to

€

a, we use the notation lim

x→ a−f ( x) . This notation is also read as “the limit of f ( x) as

€

x approaches

€

a

from the left.” To express that

€

x is decreasing to

€

a, we use the notation limx→ a+

f ( x) . This notation

is also read as “the limit of f ( x) as

€

x approaches

€

a from the right.” In either case, the actual function value at

€

x = a is not relevant. We only care about the function values as

€

x nears

€

a without

€

x ever equaling

€

a . Example 3. Consider the following piecewise-defined function:

€

f (x) = x2 if x < 2x − 4 if x > 2

Evaluate lim

x→ 2−f ( x) and lim

x→ 2+f ( x) .

What can we say about lim

x→ 2f (x )?

Solution. As

€

x increases to 2, we use the part of the function f defined for

€

x < 2 ; thus, lim

x→ 2−f ( x) = lim

x→ 2−x2 = 4.

Likewise as

€

x decreases to 2, we use the part of the function f defined for

€

x > 2, so lim

x→ 2+f ( x) = lim

x→ 2+( x − 4) = –2.

Note that the function f is not even defined at

€

x = 2, which does not matter in terms of evaluating the limits.

In order for limx→ a

f ( x) to exist,

1. Each of lim

x→ a−f ( x) and lim

x→ a+f ( x) must exist and be finite.

2. lim

x→ a−f ( x) must equal lim

x→ a+f ( x) .

When limx→ a−

f ( x) = limx→ a+

f ( x) = L , then limx→ a

f ( x) = L also.

If limx→ a−

f ( x) ≠ limx→ a+

f ( x) , then limx→ a

f ( x) does not exist.

Example 4. Let

€

f (x) =

3cos(2x) if x < π /2 6 if x = π /23sin(3x) if x > π /2.

Evaluate

€

limx→π / 2−

f (x) and

€

limx→π / 2+

f (x) . What can we say about

€

limx→π / 2

f (x)?

Solution. First, lim

x→π /2−f (x ) = lim

x→π /2−3 cos(2x ) = 3 cosπ = –3.

Next, lim

x→π /2+f (x ) = lim

x→π/ 2+3sin(3x) = 3sin(3π / 2) = –3.

Because lim

x→π /2−f (x ) = lim

x→π /2+f (x ) = –3, we can say lim

x→π /2f ( x) = –3 also.

Note that the function value when

€

x = π/2 is f (π / 2) = 6 The actual function value at this point does not affect the limit. In this case, there is a “hole in the graph,” which is also called a removable discontinuity.

Vertical Asymptotes

As mentioned earlier, a limit must be finite in order for it to be said to exist. When the value is infinite, as when approaching a vertical asymptote, then technically there is no limit since the y–values will continue to grow without bound. However we can still describe the behavior. Example 5. Let f (x) =

8

x + 4. Describe the behavior of f as

€

x approaches –4. Solution. By direct substitution of

€

x = −4 , we obtain 80

= ±∞. (For a non-zero constant

€

c , then

€

c/0 = ±∞ which means that there is a vertical asymptote.)

From the graph, we see that limx→− 4−

f (x ) = –∞ and that

limx→− 4+

f (x ) = +∞. Because these values are infinite, these

one-sided limits technically do not exist. We can still say the following: As

€

x increases to –4, then

8x + 4

decreases to –∞; and as

€

x decreases to –4, then

8x + 4

increases to +∞.

Some Basic Properties of Limits

Assume lim

x→ af ( x) and lim

x→ ag(x ) both exist. Then:

1. For any constant

€

c , we have limx→ a

c f ( x) exists and limx→ a

c f ( x) = c limx→ a

f ( x) .

That is, the limit of a constant times a function equals the constant times the limit of the function. 2. lim

x→ af (x ) ± g(x )( ) exist and lim

x→ af (x ) ± g(x )( ) = lim

x→ af ( x) ± lim

x→ ag( x) .

That is, the limit of a sum is equal to the sum of the limits, and the limit of a difference is equal to the difference of the limits. 3. lim

x→ af (x ) × g(x )( ) exists and lim

x→ af (x ) × g(x )( ) = lim

x→ af ( x) × lim

x→ag( x) .

That is, the limit of a product is equal to the product of the limits.

4. limx→ a

f ( x)g(x )

= limx→ a

f ( x)

limx→ a

g(x ), provided lim

x→ ag(x ) ≠ 0.

That is, the limit of a quotient is equal to the quotient of the limits.