The Triangle Inequality
Definitions in calculus and analysis use absolute value extensively. |x| =
x , x ≥ 0 −x, x < 0.
Fundamental properties of absolute value include |xy | = |x | · |y | and x ≤ |x |.
Another property—used often in proofs—is the triangle inequality
Definition
Sketch Proof.
Observe in the diagrams below that regardless of the order of x , y , z on the number line, the inequality |x − y | ≤ |x − z| + |z − y | holds.
x y z | {z } |x−z| |x−y | z }| { |z−y | z }| { y x z | {z } |z−y | |x−y | z }| { |x−z| z }| { z x y | {z } |z−y | |x−z| z }| { |x−y | z }| { x z y | {z } |x−y | |x−z| z }| { |z−y | z }| { y z x | {z } |x−y | |z−y | z }| { |x−z| z }| { z y x | {z } |x−z| |z−y | z }| { |x−y | z }| {
Definition of a Limit
We need to know how a certain function f (x ) behaves when x is close to some number c.
Limits are designed to deal with this type of problem.
Definition (Informal definition of a limit)
Suppose f is a function and c is a number. Then lim
x →cf (x ) = L means
that f (x ) is arbitrarily close to L provided that x is sufficiently close to c.
Definition (Precise definition of a limit)
Suppose f : X → R is a function, where X ⊆ R, and c ∈ R. Then lim
x →cf (x ) = L means that for any real ε > 0 (no matter how small), there
is a real number δ > 0 for which |f (x ) − L| < ε provided that 0 < |x − c| < δ.
Example
Prove that lim
x →2(3x + 4) = 10.
Proof.
Suppose ε > 0. Note that
|(3x + 4) − 10| = |3x − 6| = |3(x − 2)| = 3|x − 2|. So if δ = ε3 , then 0 < |x − 2| < δ yields |(3x + 4) − 10| = 3|x − 2| < 3δ = 33ε = ε. In summary, for any ε > 0, there is δ for which 0 < |x − 2| < δ implies |(3x + 4) − 10| < ε. By Definition, lim
x →2(3x + 4) = 10.
Example
Prove that lim
x →2 5x
Limits That Do Not Exist Given a function f and a number c, there are two ways that lim
x →cf (x ) = L can be false.
First, there may be a different number M 6= L for which lim
x →cf (x ) = M.
Second, it may be that Statement
∀ ε > 0, ∃ δ > 0, 0 < |x − c| < δ ⇒ |f (x) − L| < ε . (1) is false for all L ∈ R. In such a case we say that lim
Example
Suppose f (x ) = x
2 +
|x − 2|
x − 2 + 2. Prove limx →2f (x ) does not exist.
Notice that f (2) is not defined, as it involves division by zero. Also, f (x ) behaves differently depending on whether x is to the right or left of 2.
x y y = f (x ) 2 2+δ 2−δ 2 4 L+ε L L−ε a b f (a) f (b)
Suppose for the sake of contradiction that lim
x →2f (x ) = L, where L is a real
number. Let ε = 12.
By Definition, there is a real number δ > 0 for which 0 < |x − 2| < δ implies f (x ) − L < 12.
Put a = 2 −2δ , so 0 < |a − 2| < δ. Hence f (a) − L < 12. Put b = 2 +2δ , so 0 < |b − 2| < δ. Hence f (b) − L
< 12.
Further, f (a) < 2 and f (b) > 4, so 2 < |f (b) − f (a)|. With this we get a contradiction 2 < 1, as follows: 2 < f (b) − f (a) = f (b) − L − f (a) − L ≤ f (b) − L + f (a) − L < 1 2 + 1 2 = 1
Example
Prove that lim
x →0 sin
1 x
does not exist.
y = sin 1 x
x y
As x approaches 0, the number x1 grows bigger, approaching infinity, so
sin 1x just bounces up and down, faster and faster the closer x gets to 0. Intuitively, we would guess that the limit does not exist, because sin 1x does not approach any single number as x approaches 0.
Limit Laws:
Our first limit law concerns the constant function f (x ) = a where a ∈ R. Its graph is a horizontal line with y-intercept a.
Theorem (Constant function rule)
If a ∈ R, then lim
x →c a = a.
The identity function f (x ) = x .
Theorem (Identity Function Rule)
If c ∈ R, then lim
Theorem (Constant multiple rule)
If lim
x →c f (x ) exists, and a ∈ R, then limx →c af (x ) = a limx →c f (x ).
Theorem (Sum rule)
If both lim
x →c f (x ) and limx →c g (x ) exist, then
lim
Theorem (Difference rule)
If both lim
x →c f (x ) and limx →c g (x ) exist, then
lim
x →c f (x ) − g (x ) = limx →c f (x ) − limx →cg (x ).
Theorem (Multiplication rule)
If both lim lim
x →c f (x ) and limx →c g (x ) exist, then
lim x →c f (x )g (x ) = lim x →c f (x ) ·lim x →c g (x ) .
Theorem (Division Rule)
If both lim
x →cf (x ) and limx →cg (x ) exist, and limx →cg (x ) 6= 0, then
lim x →c f (x ) g (x ) = lim x →cf (x ) lim x →cg (x ) .
Example Find lim x →1 1 x − 1 1 − x.
Here x approaches 1, but simply plugging in x = 1 gives 1 1−1
1−1 = 0 0
(undefined). So we apply whatever algebra is needed to cancel the denominator 1 − x , and follow this with limit laws:
lim x →1 1 x − 1 1 − x = limx →1 1 x − 1 1 − x x x (multiply quotient by 1 = x x) = lim x →1 (1 − x ) (1 − x )x (distribute x on top) = lim x →1 1 x (cancel the (1 − x )) = lim x →11 lim x →1x = 1
