CHAPTER 1 SUPPLEMENTARY EXERCISES
2.2 LIMITS
Calculus is the mathematics of change. Sometimes, change is finite (discrete) as, for example, counting a stack of dollar bills. Other change is continuous as, for example, the trajectory and speed of a thrown ball.
For the continuous case, in which change is instantaneous, smaller and smaller intervals improve the accuracy of approximating measurements. The ball’s speed can be approxi- mated by using the time to traverse a small distance on its trajectory. The accuracy of the
LIMITS 47 approximation improves as the distance traversed decreases. The successive measurements over shorter and shorter distances represent a passage to a limit.
↑Four fundamental concepts underlie the calculus: a function, the limit, the derivative, and the integral. Functions were considered in the previous chapter, and limits and derivatives arise in this chapter; the integral is considered in Chapter 6.
Thelimitconcept is fundamental to the calculus. There are three limits that we consider in this text. Two are one-sided limits. One is a limit asxapproachesx=afrom the negative side (left) (whenx<a). A second, whenxapproachesx=afrom the positive side (right) (whenx>a). Third, when both one-sided limits approach the same number we say the limit asxapproaches a is that value. Actually,f(x) may not exist atx=a, and only the limit from the left and/or the right asxapproachesx=amay exist.
↓ To visualize these limit types consider a horizontalx-axis with an origin atx= 0 and a function,f(x), that takes real values at every point, andf(0) in particular.
Imagine a tiny bug on the left of the origin and crawling to the right toward it, never quite reachingx=0. This is the left sided limit forf(x)and denoted by lim
x→0−f(x) =f(0).
Now, imagine the tiny bug on the right side of the origin and moving toward it, again, getting closer, and most important, never actually reachingx=0. This right sided limit is denoted by lim
x→0+f(x) =f(0).
Finally, when the same limit is approached, the origin here, from both left and right sides, we simply writelim
x→0f(x) =f(0)without directional qualification. Unless otherwise indicated, limits in this text are two sided.
↑ Zeno’s Paradox, the legendary race of the tortoise and the hare puzzled thinkers for some 2000 years until the notion of limits was developed.
One-sided limits are not widely used in basic calculus courses but as an introduction to what is generally referred to as the limit. About notation: mathematicians use the symbolism limx→af(x) =f(a)to signify that the limiting value of a functionf(x) asxapproaches a,(x→a)equals the value of the function there,f(a). A similar representation applies to one-sided limits.
Example 2.2.1 One-Sided Limit
Using the following graph, determine the limits:
a) lim
x→2−f(x) b) lim
x→2+f(x) c) lim
x→4−f(x) d) lim
x→4+f(x)
–2 –1 0 1 2 3 4 5 6 7 8
–4 –2 0 4 6 8
y
x 2
Solution:
a) In this case, as x approaches 2 from the left (negative side), (to the open endpoint) the limit is 4.
b) In this case, as x approaches 2 from the right (positive side), along the horizontal, (to the closed endpoint) the limit is 3.
c) In this case, as x approaches 4 from the left (negative side), along the horizontal line, (to almost 4), the limit is 3.
d) In this case, as x approaches 4 from the right (positive side), along the horizontal line, (to almost 4), the limit is 3.
Example 2.2.2 Graphical Limit
Continuing Example 2.2.1, determine the limits a) lim
x→2f(x) b) lim
x→4f(x) c) lim
x→0f(x) d) lim
x→−1f(x)
Solution:
a) In this case, the limit does not exist as lim
x→2−f(x)≠ lim
x→2+f(x).
b) In this case, as both one-sided limits were equal to 3 the limit is 3,lim
x→4f(x) =3.
c) On either side of x=0 the function approaches 0, so this limit is 0,lim
x→0f(x) =0.
d) On either side of−1 the function approaches 1, so this limit is 1,lim
x→−1f(x) =1.
Henceforth, all limits are assumed to be two sided unless otherwise noted. Notice that there is no limit at points on the piecewise graph when there is a break (as x=2 in the previous figure). Otherwise, if there is no break in a graph near the point of interest, the limit exists there. As polynomials are continuous (unbroken) curves, a limit exists at every point. The limit is simply found by substitutingx=ainto the polynomialy=f(x).
LIMITS 49 In general, a simple way to evaluate limits ofy=f(x)asxapproaches a is to evaluate f(a). If a unique value,f(a), exists, it is the limit. If it increases indefinitely, the limit does not exist.
Occasionally, anindeterminateform, such as 0/0 or∞∕∞, is encountered. These forms require further study. Sometimes, the cause is a vanishing factor common to the numerator and denominator. Factoring and reducing fractions sometimes remove indeterminacies of the form 0/0.
Example 2.2.3 Limits Determine limits where they exist:
a) lim
x→3x4−2x3+3x+5 b) lim
x→4
x2−5x+4
x−4 c) lim
x→3
x3+2x+1 x−3 d) lim
x→2
x3−8
2x+3 e) lim
x→∞
4
3x+1 f) lim
x→∞
4x2 x2−1 Solution:
a) Substituting x=3 yields(3)4−2(3)3+3(3) +5=41 as the limit.
b) Substituting x=4 yields an indeterminate form 0/0. After factoring the numerator as (x−1)(x−4)and cancellation, the limit becomeslim
x→4(x−1) =4−1=3.
c) Substituting x=3 yields(3)3+2(3) +1
3−3 . As division by zero is undefined, this limit does not exist.
d) Substituting x=2 yields (2)3−8 2(2) +3 = 0
7 =0.
e) As x→∞, the fraction continuously decreases so, the limit is 0.
f) As x→∞, there is an indeterminate form. However, dividing numerator and denom- inator by x2and letting x→∞, the limit is 4.
Sometimes, you are given the equation of a piecewise function rather than its graph (as in Example 2.2.1). Within the specified interval, you can substitute forxto determine whether the limit exists. Endpoints of intervals may pose problems for the existence of limits. This is illustrated in the following example.
Example 2.2.4 Limits of Piecewise Functions
Determine the limits if they exist:
a) lim
x→1f(x) for f(x) = {
2x x<1 x3−1 x≥1
b) lim
x→3f(x) for f(x) =
⎧⎪
⎨⎪
⎩
4 x=3
x2−2x−3
x−3 x≠3
Solution:
a) In this case, both one-sided limits must be evaluated. If they are equal, the limit exists. To determine the limit from the negative side, evaluate the function when x is slightly less than 1 using the 2x portion of the piecewise function. Therefore,
xlim→1−f(x) =2(1) =2.
Now, evaluate the function from the right side, when x is slightly more than 1 (the x3−1 portion of the function is used). Therefore, lim
x→1+f(x) = (1)3−1=0. As the two one-sided limits differ, the limit does not exist. However, the one-sided limits exist.
b) In this case, when x is either slightly more or slightly less than 3, the second part of the piecewise function is used(x≠3). Therefore,
x→3 x→3
(x − 3)(x + 1)
lim = lim (x + 1) = 4.
(x − 3) .
EXERCISES 2.2
In Exercises 1–3, use the graph to determine indicated limits (if they exist).
–8 –6 –4 –2 0 2 4 6 8 10 12
–5 –4 –3 –2 –1 0 1 2 3 4 5 6
y
x
1. a) lim
x→3−f(x) b) lim
x→3+f(x) c)lim
x→3f(x) 2. a) lim
x→0−f(x) b) lim
x→0+f(x) c)lim
x→0f(x) 3. a) lim
x→−2−f(x) b) lim
x→−2+f(x) c) lim
x→−2f(x) In Exercises 4–6, use the graph to determine indicated limits (if they exist).
–4 –3 –2 –1 0 1 2 3 4
–4 –2 0 2 4 6
y
x
LIMITS 51 4. a) lim
x→3−f(x) b) lim
x→3+f(x) c)lim
x→3f(x) 5. a) lim
x→0−f(x) b) lim
x→0+f(x) c)lim
x→0f(x) 6. a) lim
x→−2−f(x) b) lim
x→−2+f(x) c) lim
x→−2f(x) In Exercises 7–28, determine the indicated limits if they exist.
7. lim
x→−24 8. lim
x→012 9. lim
x→5(3x+2) 10. lim
x→3(2x+7) 11. lim
x→4(9x+5) 12. lim
x→6(3x−2) 13. lim
x→1(5x2+9x+3) 14. lim
x→1(8x4+2x2+1) 15. lim
x→0(2x7+3x4+9x+3) 16. lim
x→2(4x3−6x2−7) 17. lim
x→−1(2x3+5x2−4) 18. lim
x→13+5 x
19. lim
x→2
3x x−2 20. lim
x→4
x−4 x+1 21. lim
x→3
x2−9 x−3 22. lim
x→5
x2−25 x−5 23. lim
x→2
x3−8 x−2 24. lim
x→4
x2−3x−4 2x−8 25. lim
x→∞
−4 x3 26. lim
x→∞(x2+3x+2) 27. lim
x→∞(2x3+5x+1) 28. lim
x→∞
3x2−6x+1 x2+8x+3 In Exercises 29–35, evaluate these limits for f(x), if they exist.
f(x) =
⎧⎪
⎪⎨
⎪⎪
⎩ 1
x x<0
2x+5 0≤x≤3
x2−3x−4 3<x<5
x x≥5
29. lim
x→−1f(x) 30. lim
x→0f(x) 31. lim
x→1f(x) 32. lim
x→2f(x)
33. lim
x→4f(x) 34. lim
x→5f(x) 35. lim
x→10f(x)