2. Derivatives and Their Applications - Intro to Limits

7/31/2019 2. Derivatives and Their Applications - Intro to Limits

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Differentiation/Introduction to Differentiation/Tangents, Velocity, and the Derivative

Velocity

Let f(t) denote the distance, in kilometers, a train has traveled intime t, t>0.

Let h>0. The distance the train has traveled in

the time interval from time t=t0 to time t=t0+h is

f(t0

+h)-f(t0

).

Hence thee average speed during this time interval

is (f(t0+h)-f(t0))/h.

Taking the limit as h approaches 0 gives the

speed of the train at time t=t0.

Estimate the speed of the train at time t=t0.

The speed of the train at time t= t0

is

limh0

f t0+ h( ) f t0( )

h.


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Velocity

Galileo made experiments that leadto the discovery of gravity. In the

experiments he let various objects

fall from the tower of Pisa.

The top floor of the tower (above

which the bells are hanging and

from which objects can be dropped)is about 48 meters above the

ground.

Given that the equation of motion for a freely falling

object is s=f(t)=4.9t2, compute the speed at which

a freely falling object hits the ground when it is

dropped from the top floor of the tower of Pisa.


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VelocityGiven that the equation of motion for a freely

falling object is s=f(t)=4.9t2, compute thespeed at which a freely falling object hits the

ground when it is dropped from the top floor

of the tower of Pisa.

Let us first compute the speed of the

object at time t=t0. By the previous

considerations we get:

= limh0

4.9 t0

2+ 2ht

0+ h

2( ) 4.9t02

h= lim

h0

4.9 2ht0+ h

2( )h

= limh0

4.9 2t0+ h( ) = 9.8t0.

= limh0

f t0+ h( ) f t0( )

h= lim

h0

4.9 t0+ h( )

2

4.9t0

2

hSpeed at time t0


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Velocity

Given that the equation of motion for a freelyfalling object is s=f(t)=4.9t2, compute the

speed at which a freely falling object hits the

ground when it is dropped from the top floor

of the tower of Pisa.

Height of the tower Distance fallen in time t

We know that the speed of the falling

object at time t= t0 is 9.8t0 (m/s).

To find out how long it takes for thefalling object to reach the ground,

solve t from the equation

48 = 4.9t2.

One gets t 3.13 seconds. By the previously found formula for the

speed we get:The object hits the ground with the speed of

30.7 m/s = 68.67 miles per hour.


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The Derivative and the Rate

of ChangeIn the previous examples, we computed limit of .

Writing x = x0 + h one gets

where x = x-x0is the change of the variable x, andf(x0) = f(x) f(x0) is the corresponding change in the values of

the function.


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The Derivative and the Rate

of ChangeThe limit is the derivative of thefunction f at the point x0.

limx0

fx0

( )x

This definition assumes that the limit exists. If it does, we say

that the function f is differentiable at the pointx0.

It is fairly straightforward,

that if f is differentiable

atx0, then f must be

continuous at x0. But

continuity does not

guarantee differentiability. x0

The function shown in the figure is continuous at x= x0 but notdifferentiable, because the graph of the function does not have a unique

tangent line at that point.

No unique

tangent line.


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Applications of Derivatives

Depending on the situation, the derivatives offunctions may model, for example, one of thefollowing:

1. The slope of the tangent line.

2. The speed of an object.

3. The rate at which an investment in a bankaccount grows.

4. The speed at which a hot object cools down orthe speed at which a cold object warms up.

5. Population growth or decay.


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Introduction to Limits

The Slope of a Tangent Line as a Limit

The Area of a Disk as a Limit

The Area Under the Graph of a Function

Summary


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Tangent Lines

x x + h

h f(x + h)- f(x)

Consider the problem ofdetermining the line tangentto the graph of a function f atthe point (x,f(x)).

Start by drawing a secantline that intersects the graphof the function f at thepoints (x ,f(x)) and

(x+ h,f(x + h)) (the blue linein the picture).

The slope of the blue secant ( ) ( )f f .x h xh

+

=

Letting h approach 0, the blue secant line will approach the redtangent line as the limit.

The slope of the red tangent line ist h e l i m i t of the slopes of the

secant lines as h approaches 0.

Slope of the tangent

( ) ( )0

f flimh

x h x

h

+


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Area of a Disk

To determine the area A of a disk of radiusrone can approximate such a disk withregular polygons.

To compute the area of such a regular polygonwith n sides, decompose the polygon first totriangles as indicated in the picture.

Each angle with vertex at the center is (in radians) 2/n.The polygon consists of n triangles eachhaving the area 2 sin cos .

TA r

n n

=

The area of the disk of radius r is the limitof the areas of the polygons as napproachesthe infinity.

The Area of the

Disk of Radius r

2lim sin cosn

nrn n

The familiar formula, A = r2, for the area of

a disk of radius r can be derived from this limit.

r

The picture shows anapproximation of the disk by an octagon.

cosrn

sinrn


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Estimate Areas

Consider the problem of determining the area ofthe domain bounded by the graph of the functionx2, the x-axis, and the lines x=0 and x=1.

As the number n of theapproximating rectangles grows,the approximation gets better.At t h e l i m i t we get the area of

the blue domain under thegraph of the function y = x2.This method can be applied toalmost all functions.

We determine the area by approximating thedomain with thin rectangles for which the area can be directlycomputed. Letting these rectangles get thinner, the approximationgets better and, at the limit, we get the area of the domain in question.

10

y=x2


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Limits

Understanding how functions behave as the variable approachescertain values is important for many practical applications.

Here we have discussed applications to computing the slope of a

tangent line, or to computing areas of certain domains. In physics,limits are needed, for example, in the computation of the speed ofan object.

In all of the applications presented here, the

difficulty in computing the limit is in the factthat, when inserting the limit value for thevariable, the expression evaluates to 0/0 or to0. These are undefined expressions. In

such cases, a value can be assigned to thelimit by suitable rewritings of the originalexpression. There are other methods too.

Th e l im i t o f a f u n c t i o n i s a n im p o r t a n t

cen t r a l con cep t o f ca lcu lu s .

Slope of the tangent

Inserting h = 0, weget 0/0, which isnot a number.

( ) ( )0

f flim .h

x h x

h

+


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IntroductiontoLimitsofFunctions

HeuristicDefinitionofFiniteLimitsofFunctionsExamplesofLimitComputationsInfiniteLimitsandAsymptotesSummary


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LimitsofFunctionsDefinition

Example

Notation

A function f has the finite l im i t L ata point x0 if the values f(x) getarbitrarily close to the finite numberL as xgets close to x0 but is not x0.

Observe that thevalue of f at x0has no effect onthe limit value (ifone exists). Thelimit may existeven if thefunction is notdefined forx=x0.

The function

has the limit 0 asx 0 even though

f(0) = 1.

( )

1sin , 0

f

1, 0

x xx x

x

= =

( )0

lim fx x

x L

=

Functions/Limits of Functions/I n t r oduct i on t o L im i t s o f Func t i ons


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FirstStepsinComputingLimitsTo compute the limit of a function f at a number x0, the first thingto do is to evaluate the function at x = x0. If the value of thefunction is a well defined number, then, in most cases, this value is

the actual limit.

Example Compute the limit2

1

1lim .

1xx

x

+

Solution Evaluating at x = 1, yields the value 0.2 1

1

x

x

+

We conclude that2

1

1lim 0.

1xx

x

=

+

This is, indeed, correct as can be seen from the rewriting

( ) ( )21

1 111 0.

1 1

x

x xxx

x x

+ = =

+ +



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GuessingLimitswithCalculations

Valu e o f

x

1.1 2.1

1.001 2.001

1.00001 2.000

Valu e o f

x

0.9 1.9

0.999 1.999

0.999999 2.000

In general, we may find the correct limit value by simplycalculating the values of the function near the limit point.

Example Find the value of the limit

By calculating values of near x = 1,

2

1

1lim1x

x

x

2 1

1

x

x

Solution

2 1

1

x

x

2 1

1

x

x

One concludes thatthe limit isapparently 2. Thisis, indeed, the

correct result as onecan easily show byother methods to beintroduced later.



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GuessingLimitswithCalculations

Valu e o f

x

0.1 0.4999

0.01 0.5000

0.001 0.0000

Calculators cannot, however, be always trusted.

Example Guess the value of the limit

by calculating values of near x = 0.

4

40

1 1limx

x

x

+

4

41 1x

x

+

Solution

The limit appears to be 0.This result is incorrect

4

4

1 1x

x

+

For positive values ofxsmaller than 0.001, atypical calculator gives thevalue 0 for the function.

Problems of this type arise from therounding errors that any calculatormakes in numeric computations.

Computing using limit lawsshows that

4

40

1 1 1lim .

2xx

x

+ =



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RoundingErrorsThe following graphs of the function

illustrate the rounding errors in the computation of the values ofthis expression.

( )4

41 1fx

x

x

+

=

-0.001 < x < 0.001-1 < x < 1

These graphs, produced by a computer mathematics system,illustrate the rounding error problem. The graph on the left gives acorrect idea of the behavior of the function f near x = 0. Zooming

in to the origin results to a mistake due to rounding errors.Functions/Limits of Functions/I n t r oduct i on t o L im i t s o f Func t i ons


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InfiniteLimitsDefinition

Notation

A function f has the l im i t + at a point x0 if thevalues f(x) get arbitrarily large as xgets close to x0but is not x0.

( )0

lim fx x

x

= +

Example 201

limx

x

= + This follows, since ifxis very close to0, then 1/x2 is large.

For example, ifx= 0.01,( )

22

1 110000.

0.01x= =

The fact that means that the graph of

the function has a vertical asymptote atx= 0.

20

1

limx x =

2

1

x



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InfiniteLimitsDefinition

Notation

A function f has the l i m i t at a point x0 if thevalues f(x) get arbitrarily large negative numbers asxgets close to x0 but is not x0.

( )0

lim fx x

x

=

Example 201

limx

x

x

= This follows, since ifxis very close to 0,

the (x1)/x2 is a large negative number.

For example, ifx= 0.01,( )

22

1 0.1 19000.

0.01

x

x

= =

Also in this case, the fact thatmeans that the graph of the function

has a ve r t i ca l a sym p t o t e atx= 0.

20

1limx

x

x

=

2

1x

x



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Definition

Notation

A function f has the l im i t L as xapproaches +or , if the values f(x) get arbitrarily close to thevalue L as xgets sufficiently large positive numberor sufficiently large negative number.

( )limfx

x L

=

Example 21

lim 0x

x

= This follows, since ifxis large, 1/x2

is close to 0.

For example, ifx= 1000,( )

22

1 10.000001.

1000x= =

The fact that means that the graph of

the function has the horizontal asymptote y= 0.

2

1

lim 0x x =

2

1

x

and ( )lim f .x

x L

=



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ComputingLimitsat InfinityTo compute limits at the infinity one

can use the following rules:

1. (a positive number) =

2. (a negative number) =

3. + (any finite number) =

4. (any number)/ = 0

WARNI NG. The followingare undefined:

,

0

0, /

Example2

2

1lim 1

1xx x

x

+ =

+

since

2 2

2

2

1 11

1

1111

x

x x x x

x

x

+ +

= +

+

because both 1/x and 1/x2 approach 0 as xgrows arbitrarily large.



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No Limit

The function f does nothave a limit at x=0 sincearbitrarily close to x=0 the

function f takes any valuebetween -1 and 1.

Let

f x( ) =sin 1

x

, x 0

0, x = 0

Functions/Limits of Functions/L imi t Ru les


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Summary

The limit of a function x at a number x0 is the number which thevalues of f approach as xx0.

By approach here we mean that the values of f get arbitrarily closeto the limit value as x gets sufficiently close to the limit number x

0

.

WARNING The value of the function f at x=x0 does notaffect the limit at all.

x0

L

In precise terms, gettingarbitrarily close to the limitvalue is expressed as gettingcloser than any given positivedistance .

Sufficiently close tox0 isthen expressed as theexistence of a positivenumber as in the figure.



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Limit Rules

Properties of Limits

The Squeeze Theorem

Examples


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Limits of Functions

Definition

A function f has the finite l im i t L at apoint x0 if the values f(x) get arbitrarily

close to the finite number L as xgets

close to x0 but is not x0.

This definition applies with obvious

modifications in the case of infinite limits atfinite points and in the case of limits at positiveinfinity and at the negative infinity.



27/39

Properties of LimitsAssume that , , and let

c R.

limxx

0

f x( ) = alimxx

0

g x( ) = b


limxx

0

f x( )g x( )

=

a

bprovided that b 0.

limxx

0

f x( )g x( )( ) = ablimxx

0

cf x

( )( )= ca

limxx

0

f x( )+ g x( )( ) = a+ b


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Properties of Limits

If , then exists

andlimxx

0g x( )

= a.

limxx

0

g x( )

Assume that near the number x0, but not

necessarily at x0,

f(x

)

g(x

)

h(x

).


limxx

0

f x( ) = limxx

0

h x( ) = a


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h

f

g

Squeeze Theorem Graphically

if , then exists,

and limxx

0

g x( ) = a.limxx

0

g x( )

If f(x) g(x) h(x) near x0, and


limxx

0

f x( ) = limxx

0

h x( ) = a

The values of thefunction h near thepoint x0 are squeezed

between the values ofthe functions f and g.Hence g has the samelimit as h and f.


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How to Compute Limits (1)Methods to compute limits:

1. If the function f is defined by an algebraicexpression that has finite value at the limitpoint, then this finite value is the limit.

2. If the function f is defined by an expressionwhose value is undefined at the limit point,then one either has to rewrite the expressionto a more suitable form or one has to use the

Squeeze Theorem.



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How to Compute Limits (2)


limx1

x1

1 + x2

=1 1

1 +12= 0

limx1

sin1

x

1 + cos

2

x( )

=

sin 1( )

1 + cos

2

1( )


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= limx0

x 1 + x + 1 x( )1 + x 1 x( ) 1 + x + 1 x( )

Multiply both the numerator and the denominator

by the conjugate of the denominator to get rid ofthe square roots in the denominator.


limx0

x

1 + x 1 x


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= limx0

x 1 + x + 1 x( )1 + x( )

2

1 x( )2

= limx0

x 1 + x + 1 x( )1 + x( ) 1 x( )

= limx0

x 1 + x + 1 x( )2x

= lim

x0

1 + x + 1 x( )2

= 1


limx0

x

1 + x 1 x


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limx0

xsin 1

x

For all , -1sin() 1.

Hence for allx

0.

x

xsin

1

x

x

Since , we can use the

Squeeze Theorem to conclude that

limx0

x( ) = limx0

x = 0

limx0

xsin1

x

= 0.



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MainComputationMethods

If a square root appears in the expression, then multiply anddivide by the conjugate of the square root expression.

3

( ) ( )

( ) ( )

1 2 1 21 2

1 2

1 2 30

1 2 1 2x

x x x x

x xx x

x x

x x x x

+ + + +

+ =

+ +

+

= =

+ + + + +

Cancel out common factors of rational functions.2

( ) ( )21

1 111 2.1 1 x

x xx

xx x

+

= = +

Frequently needed rule1 ( ) ( ) 2 2.a b a b a b + =

Functions/LimitsofFunctions/SolvedProblemsonLimit


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MainMethodsofLimitComputations

If the function, for which the limit needs to be computed, is

defined by an algebraic expression, which takes a finite value atthe limit point, then this finite value is the limit value.

3

If the function, for which the limit needs to be computed, cannotbe evaluated at the limit point (i.e. the value is an undefinedexpression like in (1)), then find a rewriting of the function to aform which can be evaluated at the limit point. If it is notpossible to use rewriting, use the Squeeze Theorem.

4

In the evaluation of expressions, use the rules2

( )0, , negativenumber .positive number

a = = =

The following undefined quantities cause problems:10 000 , , , ,0 , .

0



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SolvedProblemsonLimitRules


38/39

Problems

1

2

3

4

5

2

2

3 2lim

2x

x x

x

+

3 2

3 2

1lim

3 5 2x

x x x

x x x

+ + +

+ + +

2 2lim 1 1x

x x

+

2 2lim 1 1x

x x x x

+ +

2 20

2lim

2 1 3 1x

x

x x x x

+ + +



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LimitsbyRewriting1

2

2

3 2lim

2x

x x

x

+

Solution ( )( )2 1 23 2

Rewrite 1.2 2

x xx xx

x x

+= =

( )

2

2 2

3 2

Hence lim lim 1 1.2x x

x x

x

x

+

= =


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2. Derivatives and Their Applications - Intro to Limits