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This Week

Sections 2.1-2.3,2.5,2.6First homework due Tuesday night at 11:59 p.m.Average and instantaneous velocity worksheet Tuesdayavailable at http://www.math.washington.edu/∼m124/(under week 2)print it out before coming to class

Professor Christopher Hoffman Math 124

Velocity

If an arrow is shot upward on the moon with a velocity of 58meters per second. It height in meters is given by

p(t) = 58t − 0.83t2.

1 Find the average velocity over the time intervals

[1,2], [1,1.5], [1,1.1], [1,1.01], and [1,1.001].

2 Find the instantaneous velocity at t = 1.


Average Velocity between 1 and t

Average velocity is change in position divided by change intime.

Average velocity between time 1 and t isp(t)− p(1)

t − 1.

t p(t)−h(1)t−1

2 55.511.5 55.9251.1 56.257

1.01 56.33171.001 56.339


Instantaneous Velocity

To see what happens as h approaches 0 we do a little algebra.

p(1 + h)− p(1)(1 + h)− 1

=58(1 + h)− 0.83(1 + h)2 − (58− 0.83(1)2)

h

=58 + 58h − 0.83(1 + 2h + h2)− (58− 0.83)

h

=58h − 1.66h − 0.83h2

h= 56.34− 0.83h

As h approaches 0 the average velocity approaches 56.34.We say the instantaneous velocity at t = 1 is 56.34 and write

Instantaneous velocity = limh→0

p(1 + h)− p(1)h

= 56.34.


Geometric Interpretation

p(1+h)−p(1)h is the slope of the secant line between (1,p(1)) and

(1 + h,p(1 + h)).

As h approaches 0 the secant lines approach the tangent lineat (1,p(1)).

The instantaneous velocity of 56.34 is the slope of the tangentline at (1,p(1)).


Two Interpretations

All of our work over the next few weeks can be interpreted asfinding methods to determine the instantaneous velocity for anarbitrary position function.

Geometrically, our work can be interpreted as determining theslope of a tangent line for an arbitrary point on the graph of anarbitrary function.

In order to do this we first must discuss what we mean by thelimit of a function as we approach a point.


Limits

limx→a f (x) = L means that we can make the value of f (x)arbitrarily close to L by taking x sufficiently close to a but notequal to a.

In this picture limx→2 f (x) = 4.


limx→a f (x) does not depend on the value of f (a).In all of these pictures limx→0 f (x) = 1.

For most of the limits limx→a f (x) = L that we take in this coursef (a) will not be defined.


Naive Idea

We will plug in small values of x to try to find

limx→0

sin(x)x

.

x sin(x)x

±1 .84147±1/2 .95885±.4 .973545±.3 .985067±.001 .99999

Based on this we might be tempted to say limx→0 sin(x)/x = 1.


Graphical Interpretation

Looking at the graph of y = sin x/x we can see that

limx→0

sin(x)/x = 1.


Precise Definition of a Limit

The precise definition of a limit is contained in Section 2.4. Wewon’t cover this section in class. We will be content to knowlimits when we see them.


Problematic Example

Findlimx→0

sin(π/x).

x sin(π/x)±1 0±1/2 0±1/3 0±1/10 0±1/100 0±1/n 0

Based on this we might be tempted to say

limx→0

sin(π/x) = 0.


But if we would have chosen

x =2

1 + 4n

then we would have seen a different behavior as the pointsapproached 0.For these values

sin(π/x) = sin(π

2+ 2nπ

)= 1.

Also if we would have chosen

x =2

3 + 4n

then we would have sin(π/x) = sin(3π

2 + 2nπ)= −1.


sin(π/x) oscillates between −1 and 1 infinitely often as x → 0.Thus

limx→0

sin(π/x)

does not exist.


Check your work

On every limit calculation where you want to find

limx→a

f (x)

you should plug in values of x near a.

If the numbers you get are not close to your answer then youdid something wrong.


Calculating limits graphically



For the function h whose graph is given, state the value of eachquantity if it exists, If it does not exist, explain why

1 limx→−3− h(x)2 limx→−3+ h(x)3 limx→−3 h(x)4 h(−3)5 limx→0 h(x)6 h(0)7 limx→2 h(x)8 h(2)9 limx→5+ h(x)

10 limx→5− h(x)



For the function h whose graph is given, state the value of eachquantity if it exists, If it does not exist, explain why

1 limx→−3− h(x) = 42 limx→−3+ h(x) = 43 limx→−3 h(x) = 44 h(−3) does not exist5 limx→0 h(x) does not exist6 h(0) = 17 limx→2 h(x) = 28 h(2) does not exist9 limx→5+ h(x) = 3

10 limx→5− h(x) does not exist.


Infinite limits

Look at a graph of the function f (x) = ln(x).

The graph has a vertical asymptote at x = 0.As x → 0+ the function ln(x) gets more and more negative.We say

limx→0+

ln(x) = −∞.


For every n the line x = π/2 + nπ is a vertical asymptote.We can see that the one sided limits are

limx→π/2−

tan(x) =∞ and limx→π/2+

tan(x) = −∞.


The line x = a is a vertical asymptote of y = f (x) if at least oneof the following is true

limx→a f (x) =∞ or −∞limx→a+ f (x) =∞ or −∞limx→a− f (x) =∞ or −∞



For the function R whose graph is given, state the following1 limx→2 R(x)2 limx→5 R(x)3 limx→−3− R(x)4 limx→−3+ R(x)5 The equations of the vertical asymptotes.



1 limx→2 R(x) = −∞2 limx→5 R(x) =∞3 limx→−3 R(x) does not exist because

limx→−3− R(x) = −∞ while limx→−3+ R(x) =∞4 The equations of the vertical asymptotes are x = −3,

x = 2 and x = 5.


A function f is continuous at a if

limx→a

f (x) = f (a).

This definition requires three things to happen.1 f (a) exists2 limx→a+ f (x) = f (a)3 limx→a− f (x) = f (a)


Graphs of Continuous Functions

This function is continuous at every point except x = −2,2,4and 6.


Continuous Functions

Any function defined by the following functions is continuous atevery point in its domain.

polynomialsroot functionstrigonometric functionsinverse trigonometric functionsexponential functions andlogarithmic functions


Examples of Continuous Functions

All of the following functions are continuous wherever they aredefined.

f (x) = sin(

3+xx2+1

)g(t) = e3t/4+ln(2t)

r(x) =√

x3+12x−5

m(t) =√

3t + π tan−1(4 cos(t − π))


Where are the following functions continuous1 x2 sin(x)

(x−3)(x+4) is continuous except when x = 3 and x = −4

2 ln(x2 − 1) is continuous except when −1 ≤ x ≤ 13 ex/(x−1) is continuous except when x = 1.


Evaluating Limits of Continuous Functions

This is the easiest thing we will do all class all quarter.

If f (x) is continuous at a then

limx→a

f (x) = f (a).


Evaluating Limits of Continuous Functions

Find the following limits.1 limx→3

√3πx+1x2−6 =

√9π+1

3

2 limy→−2sin√

3y+105y2−3 = sin(2)

17

3 limx→−2

√3cx+127cx+1 =

√12−6cx1−14c for any c < 2 and c 6= 1/14.

When evaluating limits in this course evaluating the function atthe point should always be your first approach.

If this fails then try pugging in numbers close to a. After youmake a guess you must justify your answer.


Find

limh→0

(2 + h)3 − 8h

.

Plugging in h = 0 we get 00 .

limh→0

(2 + h)3 − 8h

= limh→0

(2 + h)3 − 8h

= limh→0

8 + 12h + 6h2 + h3 − 8h

= limh→0

12h + 6h2 + h3

h= lim

h→012 + 6h + h2

= 12.

If two functions agree everywhere except at one point a thenthe limits approaching a are the same.


Findlimt→0

1/(7 + 3t)− 1/7t

.

Plugging in t = 0 we get 00 .

limt→0

1/(7 + 3t)− 1/7t

= limt→0

7− (7 + 3t)t(7 + 3t)(7)

= limt→0

−3tt(7 + 3t)7

= limt→0

−3(7 + 3t)7

= − 349.


Rationalize the denominator

Findlim

t→16

16− t4−√

t.

Plugging in t = 16 we get 00 .

limt→16

16− t4−√

t= lim

t→16

(16− t)(4 +√

t)(4−

√t)(4 +

√t)

= limt→16

(16− t)(4 +√

t)(16− t)

= limt→16

4 +√

t

= 8


Find

limx→0

(3x− 3

x3 + x

).

Plugging in x = 0 we get∞−∞.

limx→0

(3x− 3

x3 + x

)= lim

x→0

3(x2 + 1)− 3x3 + x

= limx→0

3x2

x3 + x

= limx→0

3xx2 + 1

=3 · 0

02 + 1= 0.


Find

limx→3

x3 − 27x − 3

Plugging in x = 3 we get 00 .

limx→3

x3 − 27x − 3

= limx→0

(x − 3)(x2 + 3x + 9)x − 3

= limx→0

x2 + 3x + 9

= 27.


Limit Laws

Suppose c is a constant and the limits

limx→a

f (x) and limx→a

g(x)

exist. Then1 limx→a

[f (x) + g(x)

]= limx→a f (x) + limx→a g(x)

2 limx→a[f (x)− g(x)

]= limx→a f (x)− limx→a g(x)

3 limx→a[cf (x)

]= c limx→a f (x)

4 limx→a[f (x) · g(x)

]= limx→a f (x) · limx→a g(x)

5

limx→a

f (x)g(x)

=limx→a f (x)limx→a g(x)

if limx→a g(x) 6= 0


Find

limt→−1

(cos(

3πt + 1

)− 3)(3(t + 1)2

√−4t).

This looks like a mess. Let’s look at the graph.


Here is the graph of the function

g(t) =(cos(

3πt + 1

)− 3)(3(t + 1)2

√−4t)

along with

f (t) = −2(3(t + 1)2√−4t) and h(t) = −4(3(t + 1)2

√−4t).

Notice that as t → −1 all three functions are approaching 0.


As these are continuous functions and defined at t = −1 we get

limt→−1

−2(3(t + 1)2√−4t) = −2(3(−1 + 1)

√−4(−1) = 0

and

limt→−1

−4(3(t + 1)2√−4t) = −4(3(−1 + 1)

√−4(−1) = 0.

The function that we are interested in

g(t) =(cos(

3πt + 1

)− 3)(3(t + 1)2

√−4t)

is sandwiched in between so limt→−1 g(t) must be zero as well.


Sandwich Theorem

This theorem makes the previous slide precise.

TheoremIf f (x) ≤ g(x) ≤ h(x) in some interval around a and

limx→a

(f (x)) = limx→a

(h(x)) = L

thenlimx→a

(g(x)) = L


Findlim

x→14

x − 14√14 + x

.

First we plug in x = 14 and get

14− 14√14 + 14

=0√28

= 0.

As this is a continuous function defined at 14 we have

limx→14

x − 14√14 + x

= 0.


Limit Laws and Graphs

Here is the graph of a function f (x)

Find the following limits.1 limx→2(f (x)2)

2 limx→−2(f (x)f (x + 3))3 limx→5 f (x)(2− f (x − 3))2


Let f (x) = x−1(x+2)x2 .

1 Find all vertical asymptotes of f (x).2 Find the one sided limits of f (x) as x approaches the

asymptotes.

The function is a rational function so it is continuous at everypoint where it is defined.

It is defined everywhere the denominator is not zero, which iswhen x = 0 and x = −2.

It is important to keep signs right. We must be careful about theterms close to zero.


As x → 0+, x2 approaches 0 from the right. Also when x → 0−,x2 approaches 0 from the right.

As x → 0 the numerator x − 1 approaches −1 and x + 2approaches 2.

Thus as x → 0 from either side the numerator is negative andthe denominator is small and positive. The fraction is thus largeand negative. Because of this

limx→0

x − 1(x + 2)x2 = −∞.


As x → −2+ the numerator x − 1 approaches −3 and thedenominator approaches 0 from the positive side.

As the numerator is negative and the denominator is small andpositive the fraction is large and negative. Because of this

limx→−2+

x − 1(x + 2)x2 = −∞.

As x → −2− the numerator x − 1 approaches −3 and thedenominator approaches 0 from the negative side.

As the numerator is negative and the denominator is small andnegative the fraction is large and positive. Thus

limx→−2−

x − 1(x + 2)x2 =∞.

As the two one sided limits differ we have that

limx→−2

x − 1(x + 2)x2

does not exist.


When we graph the function we can see that there are verticalasymptotes at x = −2 and x = 0.


Definition of Limits at∞

We writelim

x→∞f (x) = L

when the values of f (x) can be made arbitrarily close to L bytaking x sufficiently large. Similarly we write

limx→−∞

f (x) = L

when the values of f (x) can be made arbitrarily close to L bytaking sufficiently large negative values for x .


Limits at∞ Graphically

Consider the function f (x) = 3− 2e−x

As x gets large f (x) gets closer to 3. The graph of f (x) getscloser to the graph of y = 3 and we write

limx→∞

3− 2e−x = 3.


Limits at∞ Graphically

As x gets more and more negative 3− 2e−x gets more andmore negative as well. We write

limx→−∞

3− 2e−x = −∞.


Horizontal Asymptotes

The line y = L is a horizontal asymptote of the curve y = f (x) ifeither

limx→∞

f (x) = L or limx→−∞

f (x) = L.

Thus y = π/2 and y = −π/2 are horizontal asymptotes and

limx→−∞

tan−1(x) = π/2 and limx→∞

tan−1(x) = −π/2


Limits of polynomials at∞

If r > 0 then limx→∞ x r =∞If r < 0 then limx→∞ x r = 0If r is even then limx→−∞ x r =∞If r is odd then limx→−∞ x r = −∞


Limits Laws at∞

Suppose c is a constant and the limits

limx→∞

f (x) and limx→∞

g(x)

exist. Then1 limx→∞

[f (x) + g(x)

]= limx→∞ f (x) + limx→∞ g(x)

2 limx→∞[f (x)− g(x)

]= limx→∞ f (x)− limx→∞ g(x)

3 limx→∞[cf (x)

]= c limx→∞ f (x)

4 limx→∞[f (x) · g(x)

]= limx→∞ f (x) · limx→∞ g(x)

5 limx→∞f (x)g(x) =

limx→∞ f (x)limx→∞ g(x) if limx→∞ g(x) 6= 0


Divide by the largest power of the denominator

Find the following limits.

limx→∞x3/2+2πx7−2x

x2−1

limx→∞−5x3+2(r−1)x√

x4+1

limx→−∞−5x3+2(r−1)x√

x4+1


limx→∞

(√x2 + 3x + 2− x

)We rationalize the numerator by multiplying the expression by

√x2 + 3x + 2 + x√x2 + 3x + 2 + x

limx→∞

(√x2 + 3x + 2− x

)= lim

x→∞

(√x2 + 3x + 2− x

) √x2 + 3x + 2 + x√x2 + 3x + 2 + x

= limx→∞

x2 + 3x + 2− x2√

x2 + 3x + 2 + x

= limx→∞

3x + 2√x2 + 3x + 2 + x


Then we multiply numerator and denominator by 1/x .

limx→∞

(√x2 + 3x + 2− x

)= lim

x→∞

(3x + 2)(1/x)(√

x2 + 3x + 2 + x)(1/x)

= limx→∞

3 + 2/x(√

1 + 3/x + 2/x2 + 1)

=limx→∞(3 + 2/x)

limx→∞(√

1 + 3/x + 2/x2 + 1)

=3

1 + 1

=32


limx→∞

(√4x4 + 5x3 + 10− (2x2 + x)

)


For all real numbers a,b and c find

limx→∞

(√ax2 + bx + c − x

)and lim

x→−∞

(√ax2 + bx + c − x

)


1 Find all vertical asymptotes of x−1(x+2)x2 .

2 Find all horizontal asymptotes of x−1(x+2)x2 .

3 Find the one sided limits of f (x) as x approaches theasymptotes.

The function is a rational function so it is continuous at everypoint where it is defined.It is defined everywhere the denominator is not zero, which iswhen x = 0 and x = −2.It is important to keep signs right. We must be careful about theterms close to zero.


As x → 0+, x2 approaches 0 from the right. Also when x → 0−,x2 approaches 0 from the right.

As x → 0 the numerator x − 1 approaches −1 and x + 2approaches 2.

Thus as x → 0 from either side the numerator is negative andthe denominator is small and positive. The fraction is thus largeand negative. Because of this

limx→0

x − 1(x + 2)x2 = −∞.


As x → −2+ the numerator x − 1 approaches −3 and thedenominator approaches 0 from the positive side.

As the numerator is negative and the denominator is small andpositive the fraction is large and negative. Because of this

limx→−2+

x − 1(x + 2)x2 = −∞.

As x → −2− the numerator x − 1 approaches −3 and thedenominator approaches 0 from the negative side.

As the numerator is negative and the denominator is small andnegative the fraction is large and positive. Thus

limx→−2−

x − 1(x + 2)x2 =∞.

As the two one sided limits differ we have that

limx→−2

x − 1(x + 2)x2

does not exist.


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