Evaluating Limits Algebraically, Exams of Mathematics

Download Evaluating Limits Algebraically and more Exams Mathematics in PDF only on Docsity!

MATH 1550 - Calculus I study

exam solution new update

(Terrie White) Louisiana State

University

M A T H 1 5 5 0 - C a l c u l u s I s t u d y e x a m s o l u t i o n ( T e r r i e W h i t e ) L S U

2 LIMITS

2.1 Limits, Rates of Change, and Tangent Lines

Preliminary Questions

1. Average velocity is defined as a ratio of which two quantities?

SOLUTION Average velocity is defined as the ratio of distance traveled to time elapsed.

2. Average velocity is equal to the slope of a secant line through two points on a graph. Which graph?

SOLUTION Average velocity is the slope of a secant line through two points on the graph of position as a function of time.

3. Can instantaneous velocity be defined as a ratio? If not, how is instantaneous velocity computed?

SOLUTION Instantaneous velocity cannot be defined as a ratio. It is defined as the limit of average velocity as time elapsed shrinks to zero.

4. What is the graphical interpretation of instantaneous velocity at a moment t = t 0?

SOLUTION Instantaneous velocity at time t = t 0 is the slope of the line tangent to the graph of position as a function of time at t = t 0.

5. What is the graphical interpretation of the following statement: The average ROC approaches the instantaneous ROC as the interval [ x 0 , x 1 ] shrinks to x 0?

SOLUTION The slope of the secant line over the interval [ x 0 , x 1 ] approaches the slope of the tangent line at x = x 0.

6. The ROC of atmospheric temperature with respect to altitude is equal to the slope of the tangent line to a graph. Which graph? What are possible units for this rate?

SOLUTION The rate of change of atmospheric temperature with respect to altitude is the slope of the line tangent to the graph of atmospheric temperature as a function of altitude. Possible units for this rate of change are ◦F / ft or ◦C / m.

Exercises

1. A ball is dropped from a state of rest at time t = 0. The distance traveled after t seconds is s(t) = 16 t^2 ft. (a) How far does the ball travel during the time interval [ 2 , 2_._ 5 ]? (b) Compute the average velocity over [ 2 , 2_._ 5 ]. (c) Compute the average velocity over time intervals [ 2 , 2_._ 01 ], [ 2 , 2_._ 005 ], [ 2 , 2_._ 001 ], [ 2 , 2_._ 00001 ]. Use this to estimate the object’s instantaneous velocity at t = 2.

SOLUTION

(a) Galileo’s formula is s(t) = 16 t^2. The ball thus travels ∆s = s( 2_._ 5 ) − s( 2 ) = 16 ( 2_._ 5 )^2 − 16 ( 2 )^2 = 36 ft. (b) The average velocity over [2 , 2_._ 5] is

∆s

s( 2_._ 5 ) − s( 2 )

= 72 ft / s_._

(c)

∆t 2_._ 5 − 2 0_._ 5

time interval [^2 ,^^2_._ 01]^ [^2 ,^^2_.^005 ]^ [^2 ,^^2.^001 ]^ [^2 ,^^2._^00001 ]

average velocity 64.16 64.08 64.016 64.

The instantaneous velocity at t = 2 is 64 ft / s.

3. Let v = 20

T as in Example 2. Estimate the instantaneous ROC of v with respect to T when T = 300 K.

SOLUTION

T interval [^300 ,^^300_.^01 ]^ [^300 ,^^300._^005 ]

average rate of change 0.577345 0.

T interval [^300 ,^^300_.^001 ]^ [^300 ,^^300._^00001 ]

42 C H A P T ER 2 LIMITS

[

]

10 8 6 4 2

0.5 1 1.5 2 2.5 3

The instantaneous rate of change is approximately 0.57735 m /( s · K ).

In Exercises 5–6, a stone is tossed in the air from ground level with an initial velocity of 15 m / s_. Its height at time t is h(t)_ = 15 t − 4_._ 9 t^2 m_._

5. Compute the stone’s average velocity over the time interval 0_._ 5 , 2_._ 5 and indicate the corresponding secant line on a sketch of the graph of h(t).

SOLUTION The average velocity is equal to

h( 2_._ 5 ) − h( 0_._ 5 ) 2 0_.^3._

The secant line is plotted with h(t) below.

h

t

7. With an initial deposit of $100, the balance in a bank account after t years is f (t) = 100 ( 1_._ 08 )t^ dollars. (a) What are the units of the ROC of f (t)? (b) Find the average ROC over [ 0 , 0_._ 5 ] and [ 0 , 1 ]. (c) Estimate the instantaneous rate of change at t = 0_._ 5 by computing the average ROC over intervals to the left and right of t = 0_._ 5.

SOLUTION (a) The units of the rate of change of f (t) are dollars / year or $ / yr. (b) The average rate of change of f (t) = 100 ( 1_._ 08 )t^ over the time interval [ t 1 , t 2 ] is given by

∆f f (t 2 ) − f (t 1 ) . ∆t t 2 − t 1

time interval [^0 ,^.^5 ]^ [^0 ,^^1 ]

average rate of change 7.8461 8

(c)

time interval [.^5 ,^.^51 ]^ [.^5 ,^.^501 ]^ [.^5 ,^.^5001 ]

average rate of change 8.0011 7.9983 7.

time interval [.^49 ,^.^5 ]^ [.^499 ,^.^5 ]^ [.^4999 ,^.^5 ]

average ROC 7.9949 7.9977 7.

The rate of change at t = 0_._ 5 is approximately $8/yr.

In Exercises 9–16, estimate the instantaneous rate of change at the point indicated.

9. P(x) = 4 x^2 − 3; x = 2 SOLUTION

x interval [^2 ,^^2_.^01 ]^ [^2 ,^^2.^001 ]^ [^2 ,^^2.^0001 ]^ [^1.^99 ,^^2 ]^ [^1.^999 ,^^2 ]^ [^1._^9999 ,^^2 ]

average rate of change 16_._ 04 16_._ 004 16_._ 0004 15_._ 96 15_._ 996 15_._ 9996

The rate of change at x = 2 is approximately 16.

11. y(x)

x + 2

; x = 2

SOLUTION

x interval [^2 ,^^2_.^01 ]^ [^2 ,^^2.^001 ]^ [^2 ,^^2.^0001 ]^ [^1.^99 ,^^2 ]^ [^1.^999 ,^^2 ]^ [^1._^9999 ,^^2 ]

average ROC −.^0623 −.^0625 −.^0625 −.^0627 −.^0625 −.^0625

Temp (F) 60 40 20 20,  20  40  60

10,000 30,

S E C T I O N 2.1 Limits, Rates of Change, and Tangent

Lines 43

The rate of change at x = 2 is approximately − 0_._ 06.

13. f (x) = ex^ ; x = 0

SOLUTION

x interval [−^0_.^01 ,^^0 ]^ [−^0.^001 ,^^0 ]^ [−^0.^0001 ,^^0 ]^ [^0 ,^^0.^01 ]^ [^0 ,^^0.^001 ]^ [^0 ,^^0._^0001 ]

average ROC 0_.^9950 0.^9995 0.^99995 1.^0050 1.^0005 1._^00005

The rate of change at x = 0 is approximately 1_._ 00.

15. f (x) = ln x ; x = 3

SOLUTION

x interval [^2_.^99 ,^^3 ]^ [^2.^999 ,^^3 ]^ [^2.^9999 ,^^3 ]^ [^3 ,^^3.^01 ]^ [^3 ,^^3.^001 ]^ [^3 ,^^3._^0001 ]

average ROC 0_.^33389 0.^33339 0.^33334 0.^33278 0.^33328 0._^33333

The rate of change at x = 3 is approximately 0_._ 333.

17. The atmospheric temperature T (in ◦F) above a certain point on earth is T = 59 − 0_._ 00356 h , where h is the altitude in feet (valid for h ≤ 37 , 000). What are the average and instantaneous rates of change of T with respect to h? Why are they the same? Sketch the graph of T for h ≤ 37 , 000.

SOLUTION The average and instantaneous rates of change of T with respect to h are both − 0_._ 00356 ◦F / ft. The rates of change are the same because T is a linear function of h with slope − 0_._ 00356.

Altitude (ft)

19. The number P(t) of E. coli cells at time t (hours) in a petri dish is plotted in Figure 9.

(a) Calculate the average ROC of P(t) over the time interval [ 1 , 3 ] and draw the corresponding secant line.

(b) Estimate the slope m of the line in Figure 9. What does m represent?

P ( t ) 10, 8, 6, 4, 2, 1, 1 2 3 t (hours) FIGURE 9 Number of E. coli cells at time t.

SOLUTION

(a) Looking at the graph, we can estimate P( 1 ) 2000 and P( 3 ) 8000. Assuming these values of P(t) , the average

rate of change is

P( 3 )^ −^ P( 1 )^

= 3000 cells / hour_._

The secant line is here:

P ( t ) 10, 8, 6, 4, 2, 1, 1 2 3 t (hours)

= − [

]

S E C T I O N 2.1 Limits, Rates of Change, and Tangent

Lines 45

(m/s) 350 300 250 200 150 100 50

50 100 150 200 250 300

0.2 0.4 0.6 0.

• (^) In graph (B), the particle is (b) speeding up and then slowing down.
• (^) In graph (C), the particle is (d) slowing down and then speeding up.
• (^) In graph (D), the particle is (a) speeding up. 25. The fraction of a city’s population infected by a flu virus is plotted as a function of time (in weeks) in Figure 14.

(a) Which quantities are represented by the slopes of lines A and B? Estimate these slopes.

(b) Is the flu spreading more rapidly at t = 1, 2, or 3?

(c) Is the flu spreading more rapidly at t = 4, 5, or 6?

Fraction infected

B

A

1 2 3 4 5 6 FIGURE 14

Weeks

SOLUTION

(a) The slope of line A is the average rate of change over the interval [ 4 , 6 ], whereas the slope of the line B is the instantaneous rate of change at t = 6. Thus, the slope of the line A ≈ ( 0_._ 28 − 0_._ 19 )/ 2 = 0_._ 045 / week, whereas the slope of the line B ≈ ( 0_._ 28 − 0_._ 15 )/ 6 = 0_._ 0217 / week. (b) Among times t 1 , 2 , 3, the flu is spreading most rapidly at t 3 since the slope is greatest at that instant; hence,

the rate of change is greatest at that instant.

(c) Among times t 4 , 5 , 6, the flu is spreading most rapidly at t 4 since the slope is greatest at that instant; hence,

the rate of change is greatest at that instant.

27. Let v 20

T as in Example 2. Is the ROC of v with respect to T greater at low temperatures or high

temperatures? Explain in terms of the graph.

SOLUTION

T (K)

As the graph progresses to the right, the graph bends progressively downward, meaning that the slope of the tangent lines

becomes smaller. This means that the ROC of v with respect to T is lower at high temperatures.

29. Sketch the graph of f (x) x( 1 x) over 0 , 1. Refer to the graph and, without making any computations,

find:

(a) The average ROC over [ 0 , 1 ] (b) The (instantaneous) ROC at x = (^12)

(c) The values of x at which the ROC is positive

SOLUTION

y

x

(a) f ( 0 ) = f ( 1 ) , so there is no change between x = 0 and x = 1

1. The average ROC is zero. (b) The tangent line to the graph of f (x) is horizontal at x = 2 ; the instantaneous ROC is zero at this point. (c) The ROC is positive at all points where the graph is rising, because the slope of the tangent line is positive at these points. This is so for all x between x = 0 and x = 0_._ 5.

[

]

[

]

46 C H A P T ER 2 LIMITS

Further Insights and Challenges

31. The height of a projectile fired in the air vertically with initial velocity 64 ft / s is h(t) = 64 t − 16 t^2 ft. (a) Compute h( 1 ). Show that h(t) − h( 1 ) can be factored with (t − 1 ) as a factor. (b) Using part (a), show that the average velocity over the interval [ 1 , t ] is − 16 (t − 3 ). (c) Use this formula to find the average velocity over several intervals [ 1 , t ] with t close to 1. Then estimate the instan- taneous velocity at time t = 1. SOLUTION

(a) With h(t) = 64 t − 16 t^2 , we have h( 1 ) = 48 ft, so

h(t) − h( 1 ) = − 16 t^2 + 64 t − 48_._

Taking out the common factor of − 16 and factoring the remaining quadratic, we get

h(t) − h( 1 ) = − 16 (t^2 − 4 t + 3 ) = − 16 (t − 1 )(t − 3 ).

(b) The average velocity over the interval [ 1 , t ] is h(t) − h( 1 ) =

− 16 (t − 1 )(t − 3 ) = − 16 (t − 3 ). t − 1 t − 1

(c)

The instantaneous velocity is approximately 32 ft / s. Plugging t 1 second into the formula in (b) yields 16 ( 1 3 ) 32 ft / s exactly.

33. Show that the average ROC of f (x) = x^3 over [ 1 , x ] is equal to x^2 + x + 1. Use this to estimate the instantaneous ROC of f (x) at x = 1. SOLUTION The average ROC is f (x) − f ( 1 ) =

x^3 − (^1).

x − 1 x − 1 Factoring the numerator as the difference of cubes means the average rate of change is

(x − 1 )(x^2 + x + 1 )

x − 1 x^

(^2) + x + 1

(for all x /= 1). The closer x gets to 1, the closer the average ROC gets to 12 + 1 + 1 = 3. The instantaneous ROC is 3.

35. Let T^3

L as in Exercise 21. The numbers in the second column of Table 4 are increasing and those in the last column are decreasing. Explain why in terms of the graph of T as a function of L. Also, explain graphically why the instantaneous ROC at L = 3 lies between 0.4329 and 0.4331.

TABLE 4 Average Rates of Change of T with Respect to L

Interval

Average ROC (^) Interval

Average ROC

[ 3 , 3_._ 2 ] (^) 0.42603 [ 2_._ 8 , 3 ] (^) 0.

[ 3 , 3_._ 1 ] 0.42946^ [ 2_._ 9 , 3 ] 0. [ 3 , 3_._ 001 ] 0.43298^ [ 2_._ 999 , 3 ] 0. [ 3 , 3_._ 0005 ] 0.43299^ [ 2_._ 9995 , 3 ] 0.

SOLUTION Since the average ROC is increasing on the intervals 3 , L as L get close to 3, we know that the slopes of the secant lines between points on the graph over these intervals are increasing. The more rows we add with smaller intervals, the greater the average ROC. This means that the instantaneous ROC is probably greater than all of the numbers in this column. Likewise, since the average ROC is decreasing on the intervals L, 3 as L gets closer to 3, we know that the slopes of the secant lines between points over these intervals are decreasing. This means that the instantaneous ROC is probably less than all the numbers in this column. The tangent slope is somewhere between the greatest value in the first column and the least value in the second column. Hence, it is between_._ 43299 and_._ 43303. The first column underestimates the instantaneous ROC by secant slopes; this estimate improves as L decreases toward L = 3. The second column overestimates the instantaneous ROC by secant slopes; this estimate improves as L increases toward L = 3.

t 1.1 1.05 1.01 1.

average velocity over [ 1 , t ]

2

5

— −

2

48 C H A P T ER 2 LIMITS

x f (x) x f (x)

1_._ 002 0_._ 998

1_._ 001 0_._ 999

1_._ 0005 0_._ 9995

1_._ 00001 0_._ 99999

SOLUTION

x 0.998 0.999 0.9995 0.99999 1.00001 1.0005 1.001 1.

f (x) (^) 1.498501 1.499250 1.499625 1.499993 1.500008 1.500375 1.500750 1.

The limit as x → 1 is 3.

3. lim y → 2

f y , where f y

y^2 −^ y −^2

y^2 + y − 6

y f (y) y f (y)

2_._ 002 1_._ 998

2_._ 001 1_._ 999

2_._ 0001 1_._ 9999

SOLUTION

y 1.998 1.999 1.9999 2.0001 2.001 2.

f (y) 0.59984 0.59992 0.599992 0.600008 0.60008 0.

The limit as y → 2 is 3.

5. lim x → 0

ex^ x 1 f (x) , where f (x) = x^2

x ±^0_.^5 ±^0.^1 ±^0.^05 ±^0._^01

f (x)

SOLUTION

x −^0_.^5 −^0.^1 −^0.^05 −^0._^01

f (x) (^) 0.426123 0.483742 0.491770 0.

x 0.01 0.05 0.1 0.

f (x) (^) 0.501671 0.508439 0.517092 0.

The limit as x → 0 is 1.

7. Determine lim x → 0_._ 5 f^ (x)^ for^ the^ function^ f^ (x)^ shown^ in^ Figure^ 9.

y

1

x

FIGURE 9

SOLUTION The graph suggests that f (x) → 1_._ 5 as x →. 5.

.

.

2

3

S E C T I O N 2.2 Limits: A Numerical and Graphical

Approach 49

In Exercises 9–10, evaluate the limit.

9. lim x x → 21

SOLUTION As x → 21, f (x) = x → 21. You can see this, for example, on the graph of f (x) = x.

In Exercises 11–20, verify each limit using the limit definition. For example, in Exercise 11, show that 2 x 6 can be

made as small as desired by taking x close to 3_._

11. lim 2 x 6 x → 3

SOLUTION | 2 x − 6 | = 2 | x − 3 |. | 2 x − 6 | can be made arbitrarily small by making x close enough to 3, thus making

| x − 3 | small.

13. lim ( 4 x 3 ) 11 x → 2

SOLUTION | ( 4 x + 3 ) − 11 | = | 4 x − 8 | = 4 | x − 2 |. Therefore, if you make | x − 2 | small enough, you can make

| ( 4 x + 3 ) − 11 | as small as desired.

15. lim ( 2 x) 18 x → 9

SOLUTION We have |− 2 x − ( − 18 ) | = | ( − 2 )(x − 9 ) | = 2 | x − 9 |. If you make | x − 9 | small enough, you can make 2 | x − 9 | = |− 2 x − ( − 18 ) | as small as desired.

17. lim x^2 x → 0

SOLUTION As x → 0, we have | x^2 − 0 | = | x + 0 || x − 0 |. To simplify things, suppose that | x | < 1, so that | x +

0 || x − 0 | = | x || x | < | x |. By making | x | sufficiently small, so that | x + 0 || x − 0 | = x^2 is even smaller, you can make

| x^2 − 0 | as small as desired.

19. lim (x^2 2 x 3 ) 3 x → 0

SOLUTION As x → 0, we have | x^2 + 2 x + 3 − 3 |= | x^2 + 2 x | = | x || x + 2 |. If | x | < 1, | x + 2 | can be no bigger

than 3, so | x || x + 2 | < 3 | x |. Therefore, by making | x − 0 |= | x | sufficiently small, you can make | x^2 + 2 x + 3 − 3 |= | x || x + 2 | as small^ as desired. In Exercises 21–36, estimate the limit numerically or state that the limit does not exist.

21. lim

x − 1

x → 1 x^ −^1

SOLUTION

x .9995 .99999 1.00001 1.

f (x) (^) .500063 .500001 .49999.

The limit as x → 1 is 1.

23. lim x^

(^2) + x − 6

x → 2 x^2 − x − 2

SOLUTION

x 1.999 1.99999 2.00001 2.

f (x) 1.666889 1.666669 1.666664 1.

The limit as x → 2 is 5.

25. lim

sin 2 x

x → 0 x

SOLUTION

x −^0_.^01 −^0._^005 0.005 0.

f (x) (^) 1.999867 1.999967 1.999967 1.

The limit as x → 0 is 2.

→

[ ] = ≤

[

]

[ x ] = x

x

x

x

x

S E C T I O N 2.2 Limits: A Numerical and Graphical

Approach 51

2

1

 (^1 )

2 4

39. The greatest integer function is defined by x n , where n is the unique integer such that n x < n 1. See

Figure 13. (a) For which values of c does (^) x lim[ x ] exist? What about x

lim [ x ]? → c − (b) For which values of c does lim x exist? x → c

→ c +

y

x

SOLUTION

FIGURE 13 Graph of y = [ x ].

(a) The one-sided limits exist for all real values of c.

(b) For each integer value of c , the one-sided limits differ. In particular, lim (^) x

[ x ] = c − 1, whereas lim x

[ x ] = c. (For

noninteger values of c , the one-sided limits both equal [ c ].) The limit lim

→ c − exists when → c +

lim x → c −

[ x ] lim → c +

x → c

[ x ] ,

namely for noninteger values of c : n < c < n + 1, where n is an integer.

In Exercises 41–43, determine the one-sided limits numerically.

41. lim x → 0 ±

sin x

| x |

SOLUTION

x −.^2 −.^02 .02.

f (x) −. 993347 −. (^999933) .999933.

The left-hand limit is lim x → 0 −

f (x) 1, whereas the right-hand limit is lim → 0 +

f (x) = 1.

43. lim x → 0 ±

x − sin ( | x | ) x^3

SOLUTION

x −.^1 −.^01 .01.

f (x) (^) 199.853 19999.8 .166666.

The left-hand limit is lim x → 0 −

f (x) , whereas the right-hand limit is lim → 0 +

f (x)

45. Determine the one-sided limits of f (x) at c = 2 and c = 4, for the function shown in Figure 15.

SOLUTION

• (^) For c 2, we have lim → 2 −
• (^) For c = 4, we have x

lim

f (x) and lim → 2 + f (x) = −∞ and x

lim

f (x) = ∞.

f (x) = 10. → 4 − → 4 +

y 15

10

5

x

 5

FIGURE 15

x

x

x

2

52 C H A P T ER 2 LIMITS

6

4

2

1234

3

2

(^1) x  1 12345

y sin 3

sin 2

y  2 x cos x x

In Exercises 47–50, draw the graph of a function with the given limits.

47. lim x → 1 f^ (x)^ 2,^ lim → 3 −

f (x) 0, lim → 3 +

f (x) = 4

SOLUTION

y

x

49. lim x → 2 +

f (x) f ( 2 ) 3, lim → 2 −

f (x) 1, lim x → 4

f (x) = 2 /= f ( 4 )

SOLUTION

y

In Exercises 51–56, graph the function and use the graph to estimate the value of the limit.

51. lim sin^3  → 0 sin 2 

SOLUTION

y

The limit as → 0 is 3.

53. lim

2 x^ − cos x x → 0 x

SOLUTION

y

The limit as x → 0 is approximately 0.693. (The exact answer is ln 2.)

55. lim → 0

cos 3 − cos 4 

^2 SOLUTION

2

54 C H A P T ER 2 LIMITS

x −.^1 −.^01 −.^001 .001 .01.

3 x^ − 1 x

We have ln 3 ≈ 1_._ 0986. bx^ − 1

• (^) We conjecture that lim x → 0 x

= ln b for any positive number b. Here are two additional test cases.

x −.^1 −.^01 −.^001 .001 .01.

1 x^ − 1 2 x

We have ln 1 ≈ − 0_._ 69315.

x −.^1 −.^01 −.^001 .001 .01.

7 x^ − 1 x

We have ln 7 ≈ 1_._ 9459.

61. Find by experimentation the positive integers k such that lim

sin ( sin^2 x)^ exists.

SOLUTION

• (^) For k 1, we have lim x → 0

f (x) lim x → 0

sin ( sin^2 x)

x

x → 0 xk

x −.^01 −.^0001 .0001.

f (x) −.^01 −.^0001 .0001.

• (^) For k 2, we have lim x → (^0) f (x) lim x → 0

sin ( sin^2 x)

x^2

x −.^01 −.^0001 .0001.

f (x) .999967 1.000000 1.000000.

• (^) For k = 3, the limit does not exist.

x −.^01 −.^0001 .0001.

f (x) (^) − 102 − 104 104 102

Indeed, as x → 0 −, f (x) = (^) sin ( sin (^2) x)

x^3

→ −∞, whereas as x → 0 +, f (x) = sin ( sin^2 x)

sin ( sin^2 x)

x^3

• (^) For k 4, we have lim x → 0

f (x) lim x → (^0) x 4 =^ ∞.

x −.^01 −.^0001 .0001.

f (x) (^) 104 108 108 104

• (^) For k = 5, the limit does not exist.

x −.^01 −.^0001 .0001.

f (x) (^) − 106 − 1012 1012 106

Indeed, as x → 0 −, f (x) = (^) sin ( sin (^2) x)

x^5

→ −∞, whereas as x → 0 +, f (x) =

sin ( sin^2 x)

x^5

S E C T I O N 2.3 Basic Limit Laws

1

 1 0. 0.  1

0.5 1

• (^) For k 6, we have lim x → (^0) f (x) lim x → 0

sin ( sin^2 x)

x^6

x −.^01 −.^0001 .0001.

f (x) (^) 108 1016 1016 108

• SUMMARY

- (^) For k 1, the limit is 0. - (^) For k 2, the limit is 1. - (^) For odd k > 2, the limit does not exist. - (^) For even k > 2, the limit is ∞. 63. The function f x

21 /x^ −^2 −^1 /x is defined for x 0. 21 /x^ + 2 −^1 /x (a) Investigate lim x → 0 +

f (x) and lim x → 0 −

f (x) numerically.

(b) Produce a graph of f on a graphing utility and describe its behavior near x = 0.

SOLUTION

(a)

x −.^3 −.^2 −.^1 .1 .2.

f (x) −.^980506 −.^998049 −.^999998 .999998 .998049.

(b) As x → 0 − , f (x) → −1, whereas as x → 0 + , f (x) → 1.

y

x

2.3 Basic Limit

Laws

Preliminary Questions

1. State the Sum Law and Quotient

Law.

FIGURE 18

SOLUTION Suppose lim x → c f (x) and lim x → c g(x) both exist. The Sum Law states that

lim ( f (x) + g(x)) = lim f (x) + lim g(x). x → c

Provided lim x → c g(x) /= 0, the Quotient Law states that

x → c x → c

lim f^ (x)^ = lim x^ → c^ f^ (x)^. x → c (^) g(x) lim x → c g(x)

2. Which of the following is a verbal version of the Product Law?

(a) The product of two functions has a limit.

(b) The limit of the product is the product of the limits.

(c) The product of a limit is a product of functions.

(d) A limit produces a product of functions.

SOLUTION The verbal version of the Product Law is (b) : The limit of the product is the product of the limits.

3. Which of the following statements are incorrect ( k and c are constants)? (a) lim k = c (b) lim k = k x → c (c) lim

x → c^2

x → 1

S E C T I O N 2.3 Basic Limit Laws

SOLUTION Statements (a) and (d) are incorrect. Because k is constant, statement (a) should read lim x → c k = k. Statement (d) should be lim x → c x = c.

4. Which of the following statements are incorrect?

(a) The Product Law does not hold if the limit of one of the functions is zero.

(b) The Quotient Law does not hold if the limit of the denominator is zero.

(c) The Quotient Law does not hold if the limit of the numerator is zero.

SOLUTION Statements (a) and (c) are incorrect. The Product Law remains valid when the limit of one or both of the

functions is zero, and the Quotient Law remains valid when the limit of the numerator is zero.

Exercises

In Exercises 1–22, evaluate the limits using the Limit Laws and the following two facts, where c and k are constants:

lim x = c, lim k = k

1. lim x x → 9

SOLUTION lim x 9. x → 9

x → c x → c

3. lim 14 x → 9

SOLUTION lim 14 14. x → 9

5. (^) x lim ( 3 x + 4 ) →− 3

SOLUTION We apply the Laws for Sums, Products, and Constants:

x

lim 3

( 3 x + 4 ) = x

lim 3

3 x + x

lim 3

→− →− →− = 3 x

lim 3

x + x

lim 3

7. (^) y lim (y + 14 )

→− →−

→− 3

SOLUTION (^) y lim (y + 14 ) = y

lim 3

y + y

lim 3

→− 3

9. lim ( 3 t 14 ) t → 4

→− →−

SOLUTION lim ( 3 t − 14 ) = 3 lim t − lim 14 = 3 · 4 − 14 = −2. t → 4

11. lim ( 4 x + 1 )( 2 x − 1 )

2

SOLUTION

t → 4 t → 4

x^ lim (^4 x^ +^1 )(^2 x^ −^1 )^ =^4 x

lim 2

x + x

lim 2

x

lim 2

x − x

lim 2

→ 1 / 2 → 1 / → 1 / → 1 / → 1 /

= 4

13. lim x(x 1 )(x 2 ) x → 2

SOLUTION We apply the Product Law and Sum Law:

lim x(x + 1 )(x + 2 ) = lim x lim (x + 1 ) lim (x + 2 ) x → 2 x → 2 x → 2 x → 2

= 2 lim x + lim 1 lim x + lim 2 x → 2 x → 2 x → 2 x → 2 = 2 ( 2 + 1 )( 2 + 2 ) = 24

15. lim

t

t → 9 t + 1

58 C H A P T ER 2 LIMITS

SOLUTION lim

t lim t =

t → 9 (^9) = =^

t → 9 t^ +^1 lim t + lim 1 9 + 1 10

17. lim

1 − x

x → 3 1 + x

t → 9 t → 9

1 − x (^) lim 1 − lim x 1 − 3 − 2 1 SOLUTION lim =

x → 3 x → 3

x → 3 1 +^ x^ lim 1 + lim x 1 + 3 4 2

19. lim t − 1

t → 2

x → 3 x → 3

SOLUTION We apply the definition of t −^1 , and then the Quotient Law.

lim t −^1 = lim lim 1 1 =

t → 2

21. lim (x^2 9 x −^3 ) x → 3

t → 2 t → 2 t^ lim^ t^^2 t → 2

SOLUTION We apply the Sum, Product, and Quotient Laws. The Product Law is applied to the exponentiations x^2 = x · x and x^3 = x · x · x.

lim (x^2

9 x

− (^3) ) = lim x (^2) + lim 9 x − 3

=

2 lim x + 9

lim

x → 3 x^ →^3 ⎛

x → 3 lim 1

⎞ x^ →^3

1

x → 3 x

28

x → 3 ( lim x)^3 9 x → 3

23. Use the Quotient Law to prove that if lim x → c (^) f (x) exists and is nonzero, then

lim

x → c (^) f (x) lim x → c

f (x)

SOLUTION Since lim x → c f^ (x)^ is^ nonzero,^ we^ can^ apply^ the^ Quotient^ Law:

lim 1 =

x^ lim → c^^1 =.

x → c f (x) (^) lim x → c

f (x)

lim x → c

f (x)

In Exercises 25–28, evaluate the limit assuming that x

lim 4

f (x) = 3 and x

lim 4

g(x) = 1_._

25. lim x →− 4

f (x)g(x)

→− →−

SOLUTION x

lim 4

f (x)g(x) = x

lim 4

f (x) x

lim 4

g(x) = 3 · 1 = 3.

27. lim x →− 4

→− g(x)

x^2

→− →−

SOLUTION Since lim x →− 4 x^

(^2) /= 0, we may apply the Quotient Law, then applying the Product Law (from x (^2) = x · x ):

g(x) (^) x lim g(x)^ lim 1 1 =

→− 4 = =. x →− 4 x^2 lim 2 x

x →− 4

lim x x →− 4

3