analysis/calculus review day 2 · 2010. 9. 15. · limits continuity diﬀerentiation integration...

Limits Continuity Differentiation Integration

Analysis/Calculus ReviewDay 2

Arvind [email protected]

Institute of Computational and Mathematical EngineeringStanford University

September 14, 2010


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


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.


Examples


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).


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.


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


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


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.


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.


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).


Tangents


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


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.


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)


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.


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.


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[.


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.


Notation

If f is Riemann integrable,

∫

A

f =

∫

A

f (x)dx = sup{L(f ,P)} = inf{U(f ,P)}


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


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).


Standard Integrals


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


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


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)


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


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


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

∣

∣

∣

∣

∣


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


r =√

x2 + y2 θ = tan−1 y

x

analysis/calculus review day 2 · 2010. 9. 15. · limits continuity diﬀerentiation integration...

Documents