mat 1320 a: calculus i - university of ottawa 16th.pdfmat 1320 a: calculus i limits and continuity...

MAT 1320 A:Calculus I

Limits andcontinuity

Derivative

MAT 1320 A: Calculus I

Paul-Eugene ParentDepartment of Mathematics and Statistics

University of Ottawa

September 16th, 2013

Derivative

Outline

1 Limits and continuity

2 Derivative

Derivative

Limits

Let f : D → R be a function, L ∈ R, and a ∈ R a point ofeither D or a boundary point of D.

Examples of acceptable “a”:

• If D = [1, 2] then a could be 1, 5;

• If D =]− 1, 3[ then a could also be 3 or −1 eventhough they are not in the domain of f .

DefinitionWe will write

limx→a

f (x) = L

if the value f (x) approaches the number L as x ∈ D goestoward a without being equal to a.

Derivative

Limits

limx→a

f (x) = L

Derivative

Limits

limx→a

f (x) = L

Derivative

Limits

limx→a

f (x) = L

Derivative Remark: That value L must be independent of the way weapproach a.

Examples: 1) What is the value of limx→xo f (x) if it exists?

Answer: 5.

Derivative2) What is the value of limx→xo f (x) if it exists?

Answer: 5.In other words what happens exactly at xo is not importantwhen computing the limit at that point.

Derivative2) What is the value of limx→xo f (x) if it exists?

Answer: 5.In other words what happens exactly at xo is not importantwhen computing the limit at that point.

Derivative

y = 1x

Derivative We can expand the definition of limits and ask: “is f (x)approaching a definite value as |x | grows arbitrarily?” Wewrite in this case either

limx→∞

f (x) or limx→−∞

f (x).

In the case of y = 1x we have

limx→∞

f (x) = limx→−∞

f (x) = 0.

Derivative We can expand the definition of limits and ask: “is f (x)approaching a definite value as |x | grows arbitrarily?” Wewrite in this case either

limx→∞

f (x) or limx→−∞

f (x).

In the case of y = 1x we have

limx→∞

f (x) = limx→−∞

f (x) = 0.

Derivative

Directional limits

In some problems we might be interested in only a certaindirection of approach.

If we want to approach “a” from the right we write

limx→a+

and from the leftlim

x→a−f (x).

Derivative

Directional limits

limx→a+

x→a−f (x).

Derivative

Directional limits

limx→a+

x→a−f (x).

Derivative

TheoremThe limit limx→a f (x) exists if and only if both the right andthe left limit exist and are equal.

Derivative

Rules when computing limits

Suppose both limx→a f (x) and limx→a g(x) are numbers andlet c ∈ R.

• limx→a

(f (x)± g(x)) = limx→a

f (x)± limx→a

• limx→a

cf (x) = c limx→a

f (x);

• limx→a

f (x)g(x) =(

limx→a

f (x))(

limx→a

• limx→a

limx→a

g(x)if lim

x→ag(x) 6= 0.

Derivative

• limx→a

f (x)± limx→a

• limx→a

cf (x) = c limx→a

f (x);

• limx→a

f (x)g(x) =(

limx→a

f (x))(

limx→a

• limx→a

limx→a

g(x)if lim

x→ag(x) 6= 0.

Derivative

• limx→a

f (x)± limx→a

• limx→a

cf (x) = c limx→a

f (x);

• limx→a

f (x)g(x) =(

limx→a

f (x))(

limx→a

• limx→a

limx→a

g(x)if lim

x→ag(x) 6= 0.

Derivative

• limx→a

f (x)± limx→a

• limx→a

cf (x) = c limx→a

f (x);

• limx→a

f (x)g(x) =(

limx→a

f (x))(

limx→a

• limx→a

limx→a

g(x)if lim

x→ag(x) 6= 0.

Derivative

Continuity

DefinitionLet f : D → R be a function and a ∈ D. We say that f iscontinuous at “a” if

f (a) = limx→a

f (x).

Derivative

Examples

For the following function

while the limx→xo f (x) exists and is equal to 5

it is notcontinuous at xo since f (xo) = 7 6= 5.

Derivative

Examples

For the following function

while the limx→xo f (x) exists and is equal to 5 it is notcontinuous at xo since f (xo) = 7 6= 5.

Derivative

Unfortunately in this case even though f (1) = 5, i.e., it isdefined at x = 1, it is not continuous since

1 = limx→1−

f (x) 6= limx→1+

f (x) = 5,

i.e., limx→1

f (x) does NOT exists!

Derivative

1 = limx→1−

f (x) 6= limx→1+

f (x) = 5,

i.e., limx→1

Derivative

1 = limx→1−

f (x) 6= limx→1+

f (x) = 5,

i.e., limx→1

Derivative

Examples of continuous functions

• All polynomials, e.g., x , x2 − 2, x3 + x − 1...

• All trigonometric functions.

• The exponential and logarithm functions.

• All rational functions, i.e., quotients of polynomial.

WARNING: One checks continuity only on x belonging tothe domain of a function.

Question: Is y = 1x continuous? YES!

Derivative

Question: Is y = 1x continuous?

Derivative

Example

Compute limx→0

(3 + x)2 − 9

Solution: First of all we notice that it is a rational functionbut 0 is not in its domain. Hence we can’t simply substitute0 in the expression. So

limx→0

(3 + x)2 − 9

x= lim

9 + 6x + x2 − 9

= limx→0

x(6 + x)

As we are interested in the limit when approaching 0 we arepurposely avoiding x = 0 hence

limx→0

(3 + x)2 − 9

x= lim

x→0(6 + x).

Derivative

Example

Compute limx→0

(3 + x)2 − 9

Solution: First of all we notice that it is a rational functionbut 0 is not in its domain. Hence we can’t simply substitute0 in the expression.

limx→0

(3 + x)2 − 9

x= lim

9 + 6x + x2 − 9

= limx→0

x(6 + x)

limx→0

(3 + x)2 − 9

x= lim

x→0(6 + x).

Derivative

Example

Compute limx→0

(3 + x)2 − 9

limx→0

(3 + x)2 − 9

x= lim

9 + 6x + x2 − 9

= limx→0

x(6 + x)

limx→0

(3 + x)2 − 9

x= lim

x→0(6 + x).

Derivative

Example

Compute limx→0

(3 + x)2 − 9

limx→0

(3 + x)2 − 9

x= lim

9 + 6x + x2 − 9

= limx→0

x(6 + x)

limx→0

(3 + x)2 − 9

x= lim

x→0(6 + x).

Derivative

Example

Compute limx→0

(3 + x)2 − 9

limx→0

(3 + x)2 − 9

x= lim

9 + 6x + x2 − 9

= limx→0

x(6 + x)

limx→0

(3 + x)2 − 9

x= lim

x→0(6 + x).

Derivative

And now we notice that the function 6 + x is a line (hencecontinuous) and that 0 is in its domain.

Conclusion: limx→0

(3 + x)2 − 9

x= lim

x→0(6 + x) = 6 + 0 = 6.

Derivative

And now we notice that the function 6 + x is a line (hencecontinuous) and that 0 is in its domain.

Conclusion: limx→0

(3 + x)2 − 9

x= lim

x→0(6 + x) = 6 + 0 = 6.

Derivative

Constructing continuous functions

Let f , g : D → R be two continuous functions (when we donot specify continuous at a particular point we meancontinuous everywhere on their domain).

• f ± g is continuous;

• f · g is continuous; and

gis continuous whenever the quotient is defined, i.e., it

is continuous at all x ∈ D such that g(x) 6= 0.

Derivative

• f · g is continuous;

Derivative

Intermediate value Theorem

Let f : D → R be a continuous function and suppose thatthere is an interval [a, b] ⊆ D.

TheoremFor each value “y” between f (a) and f (b) there isxo ∈ [a, b] such that

y = f (xo).

Derivative

Intermediate value Theorem

Let f : D → R be a continuous function and suppose thatthere is an interval [a, b] ⊆ D.

TheoremFor each value “y” between f (a) and f (b) there isxo ∈ [a, b] such that

y = f (xo).

Derivative

First application

Suppose we have a continuous function f : [0, 1]→ [0, 1].

Then one knows that there is xo ∈ [0, 1] such thatf (xo) = xo , i.e., f admits a fixed point.

Consider the new function

g : [0, 1] −→ Rx 7→ f (x)− x .

It is continuous as it is the sum of two continuous functions.

Derivative

First application

g : [0, 1] −→ Rx 7→ f (x)− x .

Derivative

First application

g : [0, 1] −→ Rx 7→ f (x)− x .

Derivative

First application

g : [0, 1] −→ Rx 7→ f (x)− x .

Derivative

First application

g : [0, 1] −→ Rx 7→ f (x)− x .

Derivative

On one hand if f (1) = 1, then we are done as we have justfound a fixed point.

Else by construction

g(1) < 0 since f (1) < 1.

On the other hand if f (0) = 0, then again we are done. Elseby construction

g(0) > 0 since f (0) > 0.

Conclusion: By the intermediate value theorem, there isxo ∈ [0, 1] such that g(xo) = 0, i.e.,

f (xo) = xo .

Derivative

On one hand if f (1) = 1, then we are done as we have justfound a fixed point. Else by construction

g(1) < 0 since f (1) < 1.

g(0) > 0 since f (0) > 0.

f (xo) = xo .

Derivative

g(1) < 0 since f (1) < 1.

On the other hand if f (0) = 0, then again we are done.

Elseby construction

g(0) > 0 since f (0) > 0.

f (xo) = xo .

Derivative

g(1) < 0 since f (1) < 1.

g(0) > 0 since f (0) > 0.

f (xo) = xo .

Derivative

g(1) < 0 since f (1) < 1.

g(0) > 0 since f (0) > 0.

f (xo) = xo .

Derivative

General culture

This result is true in higher dimensions!

It is known as Brouwer fixed point Theorem.

TheoremLet f : [0, 1]n → [0, 1]n be a continuous function forn = 1, 2, 3, ... . The there exists xo ∈ [0, 1]n such that

f (xo) = xo .

Derivative

General culture

This result is true in higher dimensions!

It is known as Brouwer fixed point Theorem.

TheoremLet f : [0, 1]n → [0, 1]n be a continuous function forn = 1, 2, 3, ... . The there exists xo ∈ [0, 1]n such that

f (xo) = xo .

Derivative

Second application

Let p(x) = xn + an−1xn−1 + ... + a1x + a0 be a polynomialof odd degree greater or equal to 1, i.e., n can be any oddpositive integer greater than or equal to one.

TheoremOne automatically knows that there is at least one xo ∈ Rsuch that

p(xo) = 0.

Intuitively we know that

limx→±∞

p(x) = limx→±∞

Derivative

Second application

p(xo) = 0.

limx→±∞

p(x) = limx→±∞

Derivative

Second application

p(xo) = 0.

limx→±∞

p(x) = limx→±∞

Derivative

Second application

p(xo) = 0.

limx→±∞

p(x) = limx→±∞

Derivative

Moreover, on one hand

limx→∞

xn =∞,

i.e., there is b ∈ R such that p(b) > 0.

On the other hand we also know since n is odd that

limx→−∞

xn = −∞,

i.e., there is a ∈ R such that p(a) < 0.

Conclusion: By the intermediate value theorem there mustexists xo ∈ [a, b] such that

p(xo) = 0.

Derivative

limx→∞

xn =∞,

limx→−∞

xn = −∞,

p(xo) = 0.

Derivative

limx→∞

xn =∞,

limx→−∞

xn = −∞,

p(xo) = 0.

Derivative

limx→∞

xn =∞,

limx→−∞

xn = −∞,

p(xo) = 0.

Derivative

limx→∞

xn =∞,

limx→−∞

xn = −∞,

p(xo) = 0.

Derivative

The derivative

Let f : D → R be a function and a ∈ D.

DefinitionWe say that f is differentiable at a ∈ D if the following limit

limh→0

f (h + a)− f (a)

is a number, i.e., it exists.

In that case we denote that number by f ′(a).

You will also see in the literaturedf

∣∣∣∣x=a

Derivative

The derivative

limh→0

f (h + a)− f (a)

∣∣∣∣x=a

Derivative

The derivative

limh→0

f (h + a)− f (a)

∣∣∣∣x=a

Derivative

The derivative

limh→0

f (h + a)− f (a)

∣∣∣∣x=a

Derivative

Let f and g two differentiable functions at “a” and c ∈ R.

• (cf )′(a) = cf ′(a);

• (f ± g)′(a) = f ′(a)± g ′(a);

• (f · g)′(a) = f ′(a) · g(a) + f (a) · g ′(a); and

)′(a) =

f ′(a) · g(a)− f (a) · g ′(a)

(g(a))2.

Derivative

• (cf )′(a) = cf ′(a);

• (f ± g)′(a) = f ′(a)± g ′(a);

• (f · g)′(a) = f ′(a) · g(a) + f (a) · g ′(a); and

)′(a) =

f ′(a) · g(a)− f (a) · g ′(a)

(g(a))2.

Derivative

• (cf )′(a) = cf ′(a);

• (f ± g)′(a) = f ′(a)± g ′(a);

• (f · g)′(a) = f ′(a) · g(a) + f (a) · g ′(a); and

)′(a) =

f ′(a) · g(a)− f (a) · g ′(a)

(g(a))2.

Derivative

• (cf )′(a) = cf ′(a);

• (f ± g)′(a) = f ′(a)± g ′(a);

• (f · g)′(a) = f ′(a) · g(a) + f (a) · g ′(a);

)′(a) =

f ′(a) · g(a)− f (a) · g ′(a)

(g(a))2.

Derivative

• (cf )′(a) = cf ′(a);

• (f ± g)′(a) = f ′(a)± g ′(a);

• (f · g)′(a) = f ′(a) · g(a) + f (a) · g ′(a); and

)′(a) =

f ′(a) · g(a)− f (a) · g ′(a)

(g(a))2.

mat 1320 a: calculus i - university of ottawa 16th.pdfmat 1320 a: calculus i limits and continuity...

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