Intuitive Definitions
(a) Limit Suppose f (x) is defined when x is near thenumber a. (This means that f is defined on some open interval that contains a, except possibly at a itself.) Then we write
x→alimf (x) = L
and say the limit of f (x), as x approaches a, equals L if we can make the values of f (x) arbitrarily close to L (as close to L as we like) by restricting x to be sufficiently close to a (on either side of a) but not equal to a.
This says that the values of f (x) approach L as x approaches a. In other words, the values of f (x) tend to get closer and closer to the number L as x gets closer and closer to the number a (from either side of a) but x 6= a.
(b) One-Sided Limit We write lim
x→a−f (x) = L ⇐⇒ lim
x<a and x→af (x)
and say the left-hand limit of f (x), as x approaches a [or the limit of f (x) as x approaches a from the left] is equal to L if we can make the values of f (x) arbitrarily close to L by restricting x to be sufficiently close to a with x less than a.
We write
lim
x→a+
f (x) = L ⇐⇒ lim
x>a and x→af (x)
and say the right-hand limit of f (x), as x approaches a [orthe limit of f (x) as x approaches a from the right] is equal to L if we can make the values of f (x) arbitrarily close to L by restricting x to be sufficiently close to a with x greater than a.
Precise Definitions
(a) Limit Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then we say that the limit of f (x) as x approaches a is L, and we write
x→alimf (x) = L
if for every number ε > 0 there is a number δ > 0 such that if 0 < |x − a| < δ then |f (x) − L| < ε.
(b) One-Sided Limit We say that
lim
x→a−f (x) = L
if for every number ε > 0 there is a number δ > 0 such that if a − δ < x < a then |f (x) − L| < ε.
We say that
lim
x→a+f (x) = L
if for every number ε > 0 there is a number δ > 0 such that if a < x < a + δ then |f (x) − L| < ε.
Limit Laws Suppose that c is a constant and the limits lim
x→af (x) and lim
x→ag(x) exist. Then 1. lim
x→af (x) + g(x) = lim
x→af (x) + lim
x→ag(x).
2. lim
x→af (x) − g(x) = lim
x→af (x) − lim
x→ag(x).
3. lim
x→ac f (x) = c lim
x→af (x).
4. lim
x→af (x)g(x) = lim
x→af (x) · lim
x→ag(x).
5. lim
x→a
f (x) g(x) =
x→alimf (x)
x→alimg(x) if lim
x→ag(x) 6= 0.
6. lim
x→af (x)n = lim
x→af (x) where n is a positive integer.
7. If lim
x→af (x) > 0, then lim
x→a
pn
f (x) = qn
x→alimf (x) where n is a positive integer.
Some Proofs
• Outline of 6. Note that there is an equality for each n ∈ N. We shall use the method of mathematical induction to prove the equality holds for each n ∈ N.
Step 1. Show that the equality holds for n = 1.
Step 2. Show that if the equality holds for n = k then the equality holds for n = k + 1.
This implies that the equality holds for each n ∈ N.
• Idea of 7. Suppose that lim
x→af (x) = L > 0. For 0 < ε < L
2, there exists δ > 0 such that if 0 < |x − a| < δ then
|f (x) − L| < ε < L 2
⇐⇒ −L
2 < f (x) − L < L 2
=⇒ 0 < L
2 < f (x) < 3L
2 for each x ∈ (a − δ, a) ∪ (a, a + δ)
=⇒ 0 < n L 2
n−1n
< f (x)n−1n + f (x)n−2n Ln1 + · · · + f (x)1nLn−2n + Ln−1n < n 3L 2
n−1n
=⇒ for each x ∈ (a − δ, a) ∪ (a, a + δ), we have
|f (x)n1 − Ln1|= |f (x) − L|
f (x)n−1n + f (x)n−2n Ln1 + · · · + f (x)1nLn−2n + Ln−1n
< ε n L
2
n−1n
where we have used the factorization
an− bn = (a − b)(an−1+ an−2b + · · · + abn−2+ bn−1) for all n ∈ N.
Remarks
(a) Theorem Let f (x) be defined when x is near a. Then
x→alimf (x) = L if and only if lim
x→a−f (x) = L = lim
x→a+f (x).
(b) We say that lim
x→af (x) 6= L if there exists an ε0 > 0 such that for every δ > 0 there exists an xδ within δ distance from a satisfying that
|xδ− a| < δ and |f (x) − L| > ε0. (c) Theorem If f (x) ≥ 0 near a and if lim
x→af (x) exists then lim
x→af (x) ≥ 0.
Proof Suppose that lim
x→af (x) = L < 0 then for 0 < ε < |L|
2 , since lim
x→af (x) = L, there exists δ > 0 such that if 0 < |x − a| < δ then
|f (x) − L| < ε = |L|
2 ⇐⇒ −|L|
2 < f (x) − L < |L|
2 =⇒ f (x) < L + |L|
2 = L −L 2 = L
2 < 0 which contradicts to that f (x) ≥ 0 near a, so lim
x→af (x) = L ≥ 0.
L/2 0 L
If f (x) ≤ g(x) when x is near a (except possibly at a) and the limits of f and g both exist as x approaches a, then
x→alimf (x) ≤ lim
x→ag(x).
Squeeze Theorem If f (x) ≤ g(x) ≤ h(x) when x is near a (except possibly at a) and
x→alimf (x) = L = lim
x→ah(x) then
x→alimg(x) = L
Examples
1. Let f (x) = c when x is near a (except possibly at a). Show that lim
x→af (x) = lim
x→ac = c.
2. Let f (x) = x when x is near a (except possibly at a). Show that lim
x→af (x) = lim
x→ax = a.
3. Show that lim
x→axn = an where n is a positive integer. [Hint: use example 2 and Limit Law 4 or 6]
4. Let a > 0. Show that lim
x→a
√n
x = √n
a where n is a positive integer.
Given ε > 0, choose δ = min(a2, nε a2n−1n
) such that if 0 < |x − a| < δ, then
x > a
2 and x
n−j n a
j−1 n >a
2
n−1n
for each 1 ≤ j ≤ n
0 a/2 a 3a/2
and
|xn1 − an1| = |x − a|
xn−1n + xn−2n an1 + · · · + xn1an−2n + an−1n
< δ n
a 2
n−1n
≤ ε
5. (Direct Substitution Property) If f is a polynomial or a rational function and a is in the domain of f, then
x→alimf (x) = f (a).
6. Note that x2− 1
x − 1 is not defined at x = 1 and
x→1lim
x2− 1
x − 1 = lim
x→1
(x − 1)(x + 1)
x − 1 = lim
x→1(x + 1) = 2.
7. Show that lim
x→0|x| = 0.
8. Show that lim
x→0sin x = 0 and lim
x→0cos x = 1.
cos x
1
1
x sin x x
tan x = sin xcos x
9. Show that lim
x→0sin1
x does not exist.
10. Show that lim
x→0x sin1 x = 0.
11. Show that lim
x→0
sin x
x = 1. [Hint: for x 6= 0, note that sin(−x)−x = sin xx and for x > 0, note that x ≥ sin x ≥ 0 and sin xcos x = tan x ≥ x]
Definition of an Infinite Limit Let f be a function defined on both sides of a, except possibly at a itself. Then lim
x→af (x) = ∞
means that the values of f (x) can he made arbitrarily large (as large as we please) by taking x sufficiently close to a, but not equal to a, i.e. for each M > 0, there is a δ > 0 such that if 0 < |x − a| < δ then f (x) > M.
and lim
x→af (x) = −∞
means that the values of f (x) can he made arbitrarily large negative (as large as we please) by taking x sufficiently close to a, but not equal to a, i.e. for each M > 0, there is a δ > 0 such that if 0 < |x − a| < δ then f (x) < −M.
Similar definitions can be given for the one-sided infinite limits
lim
x→a−f (x) = ∞ lim
x→a+f (x) = ∞ lim
x→a−f (x) = −∞ lim
x→a+f (x) = −∞
Definition The vertical line x = a is called avertical asymptoteof the curve y = f (x) if at least one of the following statements is true:
x→alimf (x) = ∞ lim
x→a−
f (x) = ∞ lim
x→a+
f (x) = ∞
x→alimf (x) = −∞ lim
x→a−f (x) = −∞ lim
x→a+f (x) = −∞
Example Note that the vertical line x = π
2 + kπ, k ∈ Z is a vertical asymptote of f (x) = tan x.
Definition A function f is continuous at a number a if
x→alimf (x) = f (a) ⇐⇒ lim
x→af (x) exists, f (a) is defined and lim
x→af (x) = f (a).
Examples
1. The following figure shows the graph of a function f. At which numbers is f discontinuous?
Why?
2. Where are each of the following functions discontinuous?
(a) f (x) = x2− x − 2 x − 2 (b) f (x) =
x2− x − 2
x − 2 if x 6= 2
1 if x = 2
(c) f (x) =
1
x2 if x 6= 0 1 if x = 0
(d) f (x) = JxK = the greatest integer ≤ x for x ∈ R.
Definitions
(a) A function f is continuous from the right at a number a if lim
x→a+
f (x) = f (a), and f is continuous from the left at a if lim
x→a−
f (x) = f (a).
Since
x→alimf (x) = f (a) ⇐⇒ lim
x→a±f (x) = f (a), f is continuous at a if and only if f is continuous from both sides of a.
(b) A function f iscontinuous on an intervalif it is continuous at every number in the interval.
Theorem If f and g are continuous at a and c is a constant, then the following functions are also continuous at a:
1. f + g 2. f − g 3. cf 4. f g 5. f
g if g(a) 6= 0
Theorems
(a) If g is continuous at a and f is continuous at g(a). then the composite function f ◦ g given by (f ◦ g)(x) = f (g(x)) is continuous at a.
(b) The Intermediate Value Theorem Suppose that f is continuous on the closed interval [a, b] and let N be any number between f (a) and f (b), where f (a) 6= f (b). Then there exists a number c in (a, b) such that f (c) = N.
Examples
(a) Let f (x) = c ∈ R for x ∈ (a, b), a constant function. Then f is continuous on (a, b).
(b) Let f (x) = x for x ∈ (a, b). Then f is continuous on (a, b).
(c) Any polynomial is continuous everywhere; that is, it is continuous on R = (−∞, ∞).
(d) Any rational function is continuous wherever it is defined; that is, it is continuous on its domain.
(e) Root functions, trigonometric functions, inverse trigonometric functions, exponential func- tions and logarithmic functions are continuous at every number in their domains.
(f) Show that there is a solution of the equation 4x3− 6x2+ 3x − 2 = 0 between 1 and 2.
Definitions
(a) Let f be a function defined on some interval (a, ∞). Then
x→∞lim f (x) = L
means that the values of f (x) can be made arbitrarily close to L by requiring x to be sufficiently large. More precisely, lim
x→∞f (x) = L if for every ε > 0 there is a corresponding number N such that
if x > N then |f (x) − L| < ε (b) Let f be a function defined on some interval (−∞, a). Then
x→−∞lim f (x) = L
means that the values of f (x) can be made arbitrarily close to L by requiring x to be sufficiently large negative. More precisely, lim
x→−∞f (x) = L if for every ε > 0 there is a corresponding number N such that
if x < N then |f (x) − L| < ε
(c) The line y = L is called a horizontal asymptote of the curve y = f (x) if either
x→∞lim f (x) = L or lim
x→−∞f (x) = L (d) Let f be a function defined on some interval (a, ∞). Then
x→∞lim f (x) = ∞
means that for every M > 0 there is a corresponding number N > 0 such that if x > N then f (x) > M
Similarly,
x→∞lim f (x) = −∞
means that for every M < 0 there is a corresponding number N > 0 such that if x > N then f (x) < M
Definitions
(a) The tangent line to the curve y = f (x) at the point P (a, f (a)) is the line through P with slope
m = lim
x→a
f (x) − f (a) x − a = lim
h→0
f (a + h) − f (a)
h provided that the limit exists.
(b) The instantaneous velocity of an object with position function f (t) at time t = a is v(a) = lim
t→a
f (t) − f (a)
t − a = lim
h→0
f (a + h) − f (a)
h provided that the limit exists.
(c) The derivative of a function f at a number a, denoted by f0(a) is f0(a) = lim
x→a
f (x) − f (a) x − a = lim
h→0
f (a + h) − f (a)
h if the limit exists.
(d) If x changes from x1 to x2, then the change in x (also called the increment of x) is ∆x = x2− x1 and the corresponding change in y is ∆y = f (x2) − f (x1). The difference quotient
∆y
∆x = f (x2) − f (x1) x2 − x1
is called the average rate of change of y with respect to x over the interval [x1, x2] and can be interpreted as the slope of the secant line joining P (x1, f (x1)) to Q(x2, f (x2)).
Definition A function f is differentiable at a if f0(a) = lim
x→a
f (x) − f (a) x − a = lim
h→0
f (a + h) − f (a)
h exists
It is differentiable on an open interval (a, b) [or (a, ∞) or (−∞, a) or (−∞, ∞)] if it is differentiable at every number in the interval.
Theorem If f is differentiable at a, then f is continuous at a.
Example Determine the interval on which the function f (x) = |x| is differentiable.
