Limits Continuity Differentiation Integration
Analysis/Calculus ReviewDay 2
Arvind [email protected]
Institute of Computational and Mathematical EngineeringStanford University
September 14, 2010
Limits Continuity Differentiation Integration
LimitDefinitionLet A ⊂ R, f : A → R and suppose that x0 is an accumulation point ofA. We say that b ∈ R is the limit of f at x0, written
limx→x0
f (x) = b
if given any ǫ > 0 there exists δ > 0 (depending on f , x0 and ǫ) such thatfor all x ∈ A, x 6= x0, |x − x0| < δ implies that |f (x)− b| < ǫ.
1. Define f : R \ {0} → R
f (x) =
{
0 x < 0
2 x > 0
{0} is an accumulation point of R \ {0} but limx→x0 f (x) does notexist.
2. Define f : R \ {0} → R
f (x) =
{
1 x 6= 0
0 x = 0
Limits Continuity Differentiation Integration
Directional Limits
Suppose that f is defined at least on ]x0, a] ⊂ R, for some a > x0. then
limx→x+
0
f (x) = b
means the limit of f with domain A =]x0, a]. In other words, for everyǫ > 0 there is a δ > 0 such that |x − x0| < δ, x > x0 impliesf (x)− b| < ǫ. Thus, we are taking the limit of f as x approaches x0 fromthe right. Similarly, we can define
limx→x−
0f (x) = b
the limit as x approaches x0 from the left. These are known as one sidedlimits.
Limits Continuity Differentiation Integration
Examples
Limits Continuity Differentiation Integration
Continuity
DefinitionLet A ⊂ R, f : A → R and let x0 ∈ A. We say that f is continuous at x0when
limx→x0
f (x) = f (x0)
This equivalent to saying that for every convergent sequence xk → x0, wehave f (xk) → f (x0).
Limits Continuity Differentiation Integration
Examples
Let f : R → R be the identity function x 7→ x . Show that f is continuous.
SolutionFix x0 ∈ R. By definition we must find δ > 0 for given ǫ > 0 such that
|x − x0| < δ implies |f (x)− f (x0)| < ǫ. This is trivially true if one
chooses δ = ǫ. Hence f is continuous.
Limits Continuity Differentiation Integration
L’Hopital’s Rule
Suppose f and g are differentiable on ]a, b[ and g ′(x) 6= 0 for x ∈]a, b[.Suppose
f ′(x)
g ′(x)→ A as x → a
If f (x) → 0 and g(x) → 0 as x → 0, or if g(x) → ∞ as x → a, then
f (x)
g(x)→ A as x → a
Example
limx→0
sin x
x= lim
x→0
d/dx(sin x)
d/dx(x)= lim
x→0
cos x
1= 1
Limits Continuity Differentiation Integration
Sequences of functions and Convergence
Definition (Pointwise Convergence)Given a sequence of functions of a set E , namely fn : E → R. Supposethat for all x ∈ E
limn→∞
fn(x) = f (x)
then we say that {fn} converges pointwise to f .
The disadvantage with pointwise convergence is even if fn are allcontinuous, f need not be continuous.For example, consider
fn(x) =
{
0 x ≥ 1k
−kx + 1 0 ≤ x < 1k
In this case, for each x ∈ [0, 1], fk(x) converges. If x 6= 0, fk(x) → 0(since fk(x) = 0 for large k), while if x = 0, fk(x) = 1. The limit is thus
f (x) =
{
0 x 6= 0
1 x = 0
Limits Continuity Differentiation Integration
Uniform Convergence
DefinitionWe say {fn} converges uniformly to f (denoted fn → f (uniformly), if forall ǫ > 0, there exists an N such that for all n ≥ N
|fn(x)− f (x)| ≤ ǫ, ∀x
For example, on R consider the following sequence,
fk(x) =
{
0 x < k
1 x ≥ k
Clearly, fk(x) → 0 pointwise. (for k large fk(x) = 0). However, fk doesnot converge to 0 uniformly, because there are points x such that|f (x)− 0|, is not small no matter how large k is.
Limits Continuity Differentiation Integration
contd.
ExampleLet
fn(x) =sin x
x, fn : R → R
Show that fn → 0 uniformly as n → ∞.
SolutionWe must show that |fn(x)− 0| = |fn(x)| gets small independent of x as
n → ∞. But fn(x) = | sin x |/n ≤ 1/n which gets small independent of x
as n → ∞.
Things to note
• Uniform convergence implies Pointwise convergence but not viceversa.
• If {fn} is a sequence of continuous functions and fn → f (uniformly),then f is a continuous function.
Limits Continuity Differentiation Integration
Differentiation
For a function f : [a, b] → R, f is called differentiable at x0 ∈]a, b[ if thelimit
f ′(x0) = limh→0
f (x0 + h)− f (x0)
h
exists. One also writes df /dx for f ′(x).
Limits Continuity Differentiation Integration
Tangents
Limits Continuity Differentiation Integration
From First Principles
Using the trigonometric identity
sin(A+ B) = sinA cosB + cosA+ sinB
let us compute the derivative of sin x
d
dxsin x = lim
h→0
sin(x + h)− sin x
h
= limh→0
sin x cos h + cos x sin h − sin x
h
= limh→0
sin x .cos h − 1
h+ cos x .
sin h
h
= sin x . limh→0
cos h − 1
h+ cos x . lim
h→0
sin h
h= 0 + cos x .1 = cos x
Limits Continuity Differentiation Integration
Continuity of Derivatives
If a function is not countinuous at x but f (x+) and f (x−) exist, then f
has a discontinuity of the first kind. This means that either
• f (x+) 6= f (x−), or
• f (x+) = f (x−) 6= f (x)
If f is not continuous as per definition but these two conditions are metwe say f has a discontinuity of the second kind.Suppose that f is differentiable on [a, b] and suppose thatf ′(a) < λ < f ′(b). Then there exists x ∈]a, b[ such that
f ′(x) = λ
If f is differentiable on [a, b] then f ′ has no simple discontinuities. It mayhave discontinuities of the second kind.
Limits Continuity Differentiation Integration
Properties of Derivatives
Let f , g be differentiable on [a, b]. Then,
• (cf )′(x) = cf ′(x)
• (f + g)′(x) = f ′(x) + g ′(x)
• Product Rule.
(f · g)′(x) = f ′(x)g(x) + f (x)g ′(x)
• Quotient Rule
(f /g)′(x) =f ′(x)
g(x)−
g ′(x)f (x)
g2(x)
• Chain Rule
h(x) = g(f (x)) ⇒ h′(x) = g ′(f (x))f ′(x)
Limits Continuity Differentiation Integration
Examples
1. Compute derivative of g(x) = sin(x2).
• Call f (x) = x2 and recall Chain Rule for g(f (x))• g ′(x) = g ′(f (x))× f ′(x) = cos(x2)× (2x)
2. Compute derivative of
f (x) =
{
x sin 1x
x 6= 0
0 x = 0
Apply Product Rule and Chain Rule
f ′(x) = sin1
x+ x
(
−1
x2
)
cos1
x= sin
1
x−
1
xcos
1
x
When x = 0 appeal to the definition
f (h)− f (0)
h= sin
1
x
The limit does not exist and therefore the derivative does not exist atx = 0.
Limits Continuity Differentiation Integration
Rolle’s Theorem
TheoremIf f : [a, b] → R is continuous, f is
differentiable on ]a, b[ and
f (a) = f (b) = 0
then there is a number c ∈]a, b[such that f ′(c) = 0.
Limits Continuity Differentiation Integration
Mean Value Theorem
TheoremIf f : [a, b] → R is continuous, f is
differentiable on ]a, b[ , there is a
point c ∈]a, b[ such that
f (b)− f (a) = f ′(c)(b − a)
CorollaryIf f is differentiable on ]a, b[ andf ′(x) = 0 for all x ∈]a, b[ then f
is a constant on ]a, b[.
Limits Continuity Differentiation Integration
Integration
Partition [a, b] as a = x0 < x1 < · · · < xn−1 < xn = b
U(f ,P) =
n−1∑
i=0
[sup{f (x) : x ∈ [xi , xi+1]}](xi+1 − xi )
L(f ,P) =n−1∑
i=0
[inf{f (x) : x ∈ [xi , xi+1]}](xi+1 − xi )
We say that f is Riemann integrable if inf{U(f ,P)} = sup{L(f ,P)} forany partition.
Limits Continuity Differentiation Integration
Notation
If f is Riemann integrable,
∫
A
f =
∫
A
f (x)dx = sup{L(f ,P)} = inf{U(f ,P)}
Limits Continuity Differentiation Integration
Properties of IntegrationFor integrable functions f1, f2, f
• f1 + f2 is integrable and cf is integrable. (Linearity)
• If f1 ≤ f2,∫ b
a
f1(x)dx ≤
∫ b
a
f2(x)dx
• Let a ≤ b ≤ c . Then,∫ c
a
f (x)dx =
∫ b
a
f (x)dx +
∫ c
b
f (x)dx
• Suppose sup |f (x)| ≤ M, then
|
∫ b
a
f (x)dx | ≤ M(b − a)
• f1 · f2 is integerable.
• If |f (x)| is integrable
|
∫ b
a
f (x)dx | ≤
∫ b
a
|f (x)|dx
Limits Continuity Differentiation Integration
Fundamental Theorem of Calculus
If f is integrable over [a, b], define
F (x) =
∫ x
a
f (t)dt
Then, F is continuous. If f is continuous at x0 ∈ (a, b) then F isdifferentiable at x0 and F ′(x0) = f (x0). F is called the anti-derivative off .
TheoremLet f : [a, b] → R be continuous. Then f has an antiderivative F and
∫ b
a
f (x)dx = F (b)− F (a)
If G is any other antiderivative of f , then we also have that∫ b
af (x)dx = G (b)− G (a).
Limits Continuity Differentiation Integration
Standard Integrals
Limits Continuity Differentiation Integration
Method of Substitution∫
f (g(x))g ′(x)dx = F (g(x)) + C
where, F is the anti-derivative of f . This follows from
d
dxF (g(x)) = F ′(g(x))g ′(x)
= f (g(x))g ′(x)
ExampleIntegrate
∫
1
1 + x2dx
Plugging in x = tan θ, dx = sec2 θdθ.
∫
1
1 + x2dx =
∫
1
sec2 θsec2 θdθ
=
∫
dθ = θ + C = tan−1 x + C
Limits Continuity Differentiation Integration
Integration by partsThe product rule for differentiation is
(f · g)′(x) = f ′(x)g(x) + f (x)g ′(x)
Thus, we can write∫
f (x)g ′(x)dx = f (x)g(x)−
∫
f ′(x)g(x)
ExampleIntegrate
∫
log xdx
We use the above formula with f (x) = log x and g ′(x) = 1. Therefore,
∫
log xdx = (log x)x −
∫
1
xxdx
= x log x − x
Limits Continuity Differentiation Integration
Differentiation under the Integral sign
Theorem (Leibniz Rule)If f is continuous on [a, b] and u(x), v(x) are differentiable functions of x
whose values lie in [a, b], then
d
dx
∫ v(x)
u(x)
f (t)dt = f (v(x))v ′(x)− f (u(x))u′(x)
Limits Continuity Differentiation Integration
Fubini’s theorem
TheoremLet A be the triangle described by a ≤ x ≤ b, c ≤ y ≤ d and let
f : A → R be continuous. Then,
∫
A
f =
∫ b
a
(
∫ d
c
f (x , y)dy
)
dx =
∫ d
c
(
∫ b
a
f (x , y)dx
)
dy
CorollaryLet φ, ψ : [a, b] → R be continuous maps such that φ(x) ≤ ψ(x) for allx ∈ [a, b] and let A = {(x , y) : a ≤ x ≤ b, φ(x) ≤ y ≤ ψ(x)}. Letf : A → R be continuous. Then,
∫
A
f =
∫ b
a
(
∫ ψ(x)
φ(x)
f (x , y)dy
)
dx
Limits Continuity Differentiation Integration
Example
For the region R in the figure
∫
R
sin x
xdA
∫ 1
0
(∫ x
0
sin x
xdy
)
dx =
∫ 1
0
(
ysin x
x|y=xy=0
)
dx
=
∫ 1
0
sin x dx
= − cos(1) + 1
whereas, we can’t compute
∫ 1
0
∫ 1
y
sin x
xdydx
Limits Continuity Differentiation Integration
Change of Variables
TheoremLet A be an open bounded set with volume and let g : A → R
n be a C1
mapping which is one-to-one. Let B = g(A) have volume. For
f : B → R bounded and integrable, f ◦ g |J(g)| is integrable on A and
∫
B
f (y1, . . . , yn)dy1 . . . dyn =
∫
A
f (g(x1, . . . , xn))∂(g1, . . . , gn)
∂(x1, . . . , xn)dx1 . . . dxn
In addition, we require Jg(x) 6= 0 for all x ∈ A and |Jg(x)|, 1/|Jg(x)| arebounded on A. For example,
∂(f1, f2)
∂(x , y)=
∣
∣
∣
∣
∣
∂f1∂x
∂f1∂y
∂f2∂x
∂f2∂y
∣
∣
∣
∣
∣
Limits Continuity Differentiation Integration
Polar Coordinates
Define the mapping g(r , θ) : (x , y) → (r cos θ, r sin θ)
Jg(r , θ) =
∣
∣
∣
∣
cos θ −r sin θsin θ r cos θ
∣
∣
∣
∣
= r cos2 θ + r sin2 θ = r
Limits Continuity Differentiation Integration
r =√
x2 + y2 θ = tan−1 y
x