Math 112 Lecture 1: Introduction to Limits - Okanagan...

Defining Limits One-Sided Limits Infinite Limits

Lecture 1: Introduction to Limits

Defining LimitsWhy Limits?Example 1 – An Indeterminate FormDefinition of LimitExample 2 – Applying the Definition

One-Sided LimitsDoes a Limit Always Exist?Example 3 – A Step FunctionDefinition of One-Sided LimitsChecking for the Existence of a LimitExample 4 – Calculating Limits

Infinite LimitsDefinition of Infinite LimitsExample 5 – Evaluating Infinite Limits

Clint Lee Math 112 Lecture 1: Introduction to Limits 1/22


Why Limits?

Why Limits?

Limits are used in calculus to



Why Limits?

Why Limits?


define and calculate the slope of a tangent line



Why Limits?

Why Limits?



calculate velocities and rates of change



Why Limits?

Why Limits?




define and calculate areas and volumes



Why Limits?

Why Limits?




define and calculate areas and volumes

But, more generally, limits provide a way to extend the operation offunction evaluation.



Example 1 – An Indeterminate Form


Consider the function

f (x) =x + 1

x2 − 1

The domain of this function is






f (x) =x + 1

x2 − 1

The domain of this function is{

x∣

∣ − ∞ < x < −1 or − 1 < x < 1 or 1 < x < ∞

}

=






f (x) =x + 1

x2 − 1


x∣

∣ − ∞ < x < −1 or − 1 < x < 1 or 1 < x < ∞

}

=

(−∞,−1) ∪ (−1, 1) ∪ (1, ∞)






f (x) =x + 1

x2 − 1


x∣

∣ − ∞ < x < −1 or − 1 < x < 1 or 1 < x < ∞

}

=

(−∞,−1) ∪ (−1, 1) ∪ (1, ∞)

That is the function f is defined except at x = 1 and x = −1. Let’s seewhat happens when we evaluate f at these points.




Continuing Example 1

Evaluating f at x = 1 gives

f (1) =2

0

which is obviously undefined. Use yourgraphing calculator or Maple to plot thegraph of this function in the viewingrectangle

[

−2, 3]

×[

−4, 4]

.






f (1) =2

0


[

−2, 3]

×[

−4, 4]

.

1

Does your graph look like this?






f (1) =2

0


[

−2, 3]

×[

−4, 4]

.

1

Does your graph look like this? What’s wrong with this picture?






f (1) =2

0


[

−2, 3]

×[

−4, 4]

.

1

The almost vertical line through x = 1 is an artifact. The function is notdefined at x = 1, or at x = −1.






f (1) =2

0


[

−2, 3]

×[

−4, 4]

.

x=

1

Here is the correct graph with a vertical asymptote at x = 1 and an opencircle at x = −1.





Evaluating f at x = −1 gives

f (−1) =0

0

which is again undefined. However, now we do not necessarily getan arbitrarily large value, as in the last case.






f (−1) =0

0

which is again undefined. However, now we do not necessarily getan arbitrarily large value, as in the last case. This is an indeterminateform.






f (−1) =0

0

which is again undefined. However, now we do not necessarily getan arbitrarily large value, as in the last case. This is an indeterminateform. Something different happens at x = −1 than at x = 1.






f (−1) =0

0

which is again undefined. However, now we do not necessarily getan arbitrarily large value, as in the last case. This is an indeterminateform. Something different happens at x = −1 than at x = 1. Thedifference is hinted at by the graph we saw earlier.






f (−1) =0

0

which is again undefined. However, now we do not necessarily getan arbitrarily large value, as in the last case. This is an indeterminateform. Something different happens at x = −1 than at x = 1. Thedifference is hinted at by the graph we saw earlier. There is an opencircle on the graph at x = −1.





To investigate what happens to fat x = −1 make a table of valuesnear x = −1.

x f (x)






x f (x)

−2.000 −0.333333

0.000 −1.000000






x f (x)

−2.000 −0.333333

−1.500 −0.400000

−0.500 −0.666667

0.000 −1.000000






x f (x)

−2.000 −0.333333

−1.500 −0.400000

−1.100 −0.476190

−0.900 −0.526316

−0.500 −0.666667

0.000 −1.000000






x f (x)

−2.000 −0.333333

−1.500 −0.400000

−1.100 −0.476190

−1.010 −0.497512

−0.990 −0.502513

−0.900 −0.526316

−0.500 −0.666667

0.000 −1.000000






x f (x)

−2.000 −0.333333

−1.500 −0.400000

−1.100 −0.476190

−1.010 −0.497512

−1.001 −0.499750

−0.999 −0.500250

−0.990 −0.502513

−0.900 −0.526316

−0.500 −0.666667

0.000 −1.000000






It appears that the value of f getsclose to

−0.5 = −1

2

as x gets close to −1.

x f (x)

−2.000 −0.333333

−1.500 −0.400000

−1.100 −0.476190

−1.010 −0.497512

−1.001 −0.499750

−0.999 −0.500250

−0.990 −0.502513

−0.900 −0.526316

−0.500 −0.666667

0.000 −1.000000



Definition of Limit

Definition of Limit

From Example 1 we see that even though f (−1) is not defined there isa finite and precisely defined limiting value for f at x = −1.



Definition of Limit

Definition of Limit

From Example 1 we see that even though f (−1) is not defined there isa finite and precisely defined limiting value for f at x = −1. We saythat

limx→−1

x + 1

x2 − 1= −

1

2

The following definition makes this more precise.



Definition of Limit

Definition of Limit

From Example 1 we see that even though f (−1) is not defined there isa finite and precisely defined limiting value for f at x = −1. We saythat

limx→−1

x + 1

x2 − 1= −

1

2

The following definition makes this more precise.

Definition (Limit of a Function)

We say thatlimx→a

f (x) = L

if f (x) can be made arbitrarily close to L by making x sufficiently closeto a, on either side of a.



Example 2 – Applying the Definition


To apply the definition, suppose thatwe want to make sure that the valueof f (x) is within 0.2 of limit valueL = −0.5. This means that we want

−0.7 ≤ f (x) ≤ −0.3






−0.7 ≤ f (x) ≤ −0.3

From the table of values in Example 1 we see that

f (−1.5) = −0.40000 < −0.3 and f (−0.5) = −0.666667 > −0.7






−0.7 ≤ f (x) ≤ −0.3


f (−1.5) = −0.40000 < −0.3 and f (−0.5) = −0.666667 > −0.7

Hence, we only need to have x within 0.5 of the limit point x = −1, sothat

− 1.5 ≤ x ≤ −0.5






−0.7 ≤ f (x) ≤ −0.3


f (−1.5) = −0.40000 < −0.3 and f (−0.5) = −0.666667 > −0.7


− 1.5 ≤ x ≤ −0.5

Hence for x in the interval [−1.5,−0.5] the values of f (x) are in theinterval [−0.6666667,−0.4], which is contained in the interval[−0.7,−0.3].






−0.7 ≤ f (x) ≤ −0.3

The graph shows the relationbetween these two intervals.

−1.5 −0.5

−0.3

−0.7


f (−1.5) = −0.40000 < −0.3 and f (−0.5) = −0.666667 > −0.7


− 1.5 ≤ x ≤ −0.5

Hence for x in the interval [−1.5,−0.5] the values of f (x) are in theinterval [−0.6666667,−0.4], which is contained in the interval[−0.7,−0.3]. The graph of the function f lies entirely in the boxbounded by the horizontal lines and the vertical lines.





If we decrease the vertical extent of the box, which means we want tokeep the values of f (x) closer to the limit value, we must decrease itshorizontal extent as well. From the graph it appears that no matterhow small we make the vertical extent of the box, we can make thebox narrow enough (horizontally) so that the graph of f stays entirelyinside the box.





If we decrease the vertical extent of the box, which means we want tokeep the values of f (x) closer to the limit value, we must decrease itshorizontal extent as well. From the graph it appears that no matterhow small we make the vertical extent of the box, we can make thebox narrow enough (horizontally) so that the graph of f stays entirelyinside the box.To see another step in the process of further reducing the verticalextent, width, of the box click on the button below.





If we decrease the vertical extent of the box, which means we want tokeep the values of f (x) closer to the limit value, we must decrease itshorizontal extent as well. From the graph it appears that no matterhow small we make the vertical extent of the box, we can make thebox narrow enough (horizontally) so that the graph of f stays entirelyinside the box.To see another step in the process of further reducing the verticalextent, width, of the box click on the button below.

Reducing width of box



Does a Limit Always Exist?


An important feature of the definition of the limit of a function is






Two Sided Nature of Limits

We must be able to make f (x) arbitrarily close to L by making xsufficiently close to a, on either side of a.








In some cases we get a different value if we make x close to a ondifferent sides of a.








In some cases we get a different value if we make x close to a ondifferent sides of a. In this case the limit does not exist.








In some cases we get a different value if we make x close to a ondifferent sides of a. In this case the limit does not exist.This leads to the idea of one-sided limits.








In some cases we get a different value if we make x close to a ondifferent sides of a. In this case the limit does not exist.This leads to the idea of one-sided limits. We will introduce afunction in which one-sided limits play a role in Example 3 and thengive the definition of one-sided limits.








In some cases we get a different value if we make x close to a ondifferent sides of a. In this case the limit does not exist.This leads to the idea of one-sided limits. We will introduce afunction in which one-sided limits play a role in Example 3 and thengive the definition of one-sided limits.There are other cases in which a limit does not exist. We willinvestigate some of them later.



Example 3 – A Step Function



g(x) =|x|

x

−1

1

x

y






g(x) =|x|

x=

{

−1 if x < 01 if x > 0

This is a piecewise function forwhich the formula for the functionis different on different intervals.

−1

1

x

y






g(x) =|x|

x=

{

−1 if x < 01 if x > 0

This is a piecewise function forwhich the formula for the functionis different on different intervals.Its graph looks like this:

−1

1

x

y






g(x) =|x|

x=

{

−1 if x < 01 if x > 0

This is a piecewise function forwhich the formula for the functionis different on different intervals.Its graph looks like this:

−1

1

x

y

This particular piecewise functionis called a step function.




The Absolute Value Function (A digression)

The absolute value function isanother piecewise function. Itsformula is

x

y






|x| =

−x if x < 00 if x = 0x if x > 0

x

y






|x| =

−x if x < 00 if x = 0x if x > 0

Its graph looks like this:

x

y






|x| =

−x if x < 00 if x = 0x if x > 0

Its graph looks like this:

x

y

There is no jump in the graph ofthis piecewise function. Not allpiecewise functions have a jumpin their graphs.





For the function g defined at the beginning of Example 3, what canwe say about the limit

limx→0

g(x)?






limx→0

g(x)?

We must say that the limit does not exist. This is because we cannotmake g(x) close to a single value as x gets close to x = 0.






limx→0

g(x)?

We must say that the limit does not exist. This is because we cannotmake g(x) close to a single value as x gets close to x = 0. In fact, forall x < 0, g(x) = −1, and for all x > 0, g(x) = 1.






limx→0

g(x)?

We must say that the limit does not exist. This is because we cannotmake g(x) close to a single value as x gets close to x = 0. In fact, forall x < 0, g(x) = −1, and for all x > 0, g(x) = 1.However, we can write

limx→0−

g(x) = −1 and limx→0+

g(x) = 1

These are one-sided limits.



Definition of One-Sided Limits


As seen in Example 3 a function can get close to different values asthe limit point a is approached from the two different sides. Thisleads to:






Definition (Limits from the Right and Left)







We say thatlim

x→a+f (x) = R

if f (x) can be made arbitrarily close to R by making x sufficientlyclose to a with x > a, that is for x to the right of a. This is the limit of fat a from the right, or from above,







We say thatlim

x→a+f (x) = R

if f (x) can be made arbitrarily close to R by making x sufficientlyclose to a with x > a, that is for x to the right of a. This is the limit of fat a from the right, or from above, and

limx→a−

f (x) = L

if f (x) can be made arbitrarily close to L by making x sufficiently closeto a with x < a, that is for x to the left of a. This is the limit of f at afrom the left, or from below.



Checking for the Existence of a Limit


We can check to see if a limit exists by checking the one-sided limits.






Existence of a Limit







The limitlimx→a

f (x)

exists and is equal to L if and only if both







The limitlimx→a

f (x)


limx→a−

f (x) and limx→a+

f (x)

exist, and







The limitlimx→a

f (x)


limx→a−

f (x) and limx→a+

f (x)

exist, andlim

x→a−f (x) = L and lim

x→a+f (x) = L







The limitlimx→a

f (x)


limx→a−

f (x) and limx→a+

f (x)

exist, andlim

x→a−f (x) = L and lim

x→a+f (x) = L

Otherwise, the regular, two-sided, limit does not exist.



Example 4 – Calculating Limits



h(x) =

2x + 8 if x < −1−2x + 4 if −1 ≤ x < 1

x2 − 2x + 2 if x ≥ 1

Graph each branch separately:2

4

6

2−2x

y






h(x) =

2x + 8 if x < −1−2x + 4 if −1 ≤ x < 1

x2 − 2x + 2 if x ≥ 1

Graph each branch separately:

y = 2x + 8

line thru (−3, 2) and (−1, 6)

2

4

6

2−2x

y






h(x) =

2x + 8 if x < −1−2x + 4 if −1 ≤ x < 1

x2 − 2x + 2 if x ≥ 1


y = 2x + 8

line thru (−3, 2) and (−1, 6)

y = −2x + 4

line thru (−1, 6) and (1, 2)

2

4

6

2−2x

y






h(x) =

2x + 8 if x < −1−2x + 4 if −1 ≤ x < 1

x2 − 2x + 2 if x ≥ 1


y = 2x + 8

line thru (−3, 2) and (−1, 6)

y = −2x + 4

line thru (−1, 6) and (1, 2)

y = x2 − 2x + 2 = (x − 1)2 + 1

parabola

2

4

6

2−2x

y






h(x) =

2x + 8 if x < −1−2x + 4 if −1 ≤ x < 1

x2 − 2x + 2 if x ≥ 1


y = 2x + 8

line thru (−3, 2) and (−1, 6)

y = −2x + 4

line thru (−1, 6) and (1, 2)

y = x2 − 2x + 2 = (x − 1)2 + 1

parabola

2

4

6

2−2x

y

This is a piecewise functionwith a jump at x = 1 and acorner at x = −1.





Calculate each limit, if it exists. If the limit does not exist, explainwhy.

(a) (b)

(c) (d)

(e) (f)






(a) limx→−1−

h(x) (b) limx→−1+

h(x)

(c) limx→−1

h(x) (d) limx→1−

h(x)

(e) limx→1+

h(x) (f) limx→1

h(x)






(a) limx→−1−

h(x) = 6 (b) limx→−1+

h(x)

(c) limx→−1


h(x)

(e) limx→1+

h(x) (f) limx→1

h(x)






(a) limx→−1−

h(x) = 6 (b) limx→−1+

h(x) = 6

(c) limx→−1


h(x)

(e) limx→1+

h(x) (f) limx→1

h(x)






(a) limx→−1−

h(x) = 6 (b) limx→−1+

h(x) = 6

(c) limx→−1

h(x) = 6 (d) limx→1−

h(x)

(e) limx→1+

h(x) (f) limx→1

h(x)






(a) limx→−1−

h(x) = 6 (b) limx→−1+

h(x) = 6

(c) limx→−1

h(x) = 6 (d) limx→1−

h(x) = 2

(e) limx→1+

h(x) (f) limx→1

h(x)






(a) limx→−1−

h(x) = 6 (b) limx→−1+

h(x) = 6

(c) limx→−1

h(x) = 6 (d) limx→1−

h(x) = 2

(e) limx→1+

h(x) = 1 (f) limx→1

h(x)






(a) limx→−1−

h(x) = 6 (b) limx→−1+

h(x) = 6

(c) limx→−1

h(x) = 6 (d) limx→1−

h(x) = 2

(e) limx→1+

h(x) = 1 (f) limx→1

h(x) DNE






(a) limx→−1−

h(x) = 6 (b) limx→−1+

h(x) = 6

(c) limx→−1

h(x) = 6 (d) limx→1−

h(x) = 2

(e) limx→1+

h(x) = 1 (f) limx→1

h(x) DNE

The two-sided limit at x = 1 does not exist (DNE) because the twoone-sided limits at x = 1 are different.



Infinite Limits


f (x) =x + 1

x2 − 1

from Example 1. This function takes arbitrarily large, positive andnegative, values near x = 1. So the function is undefined there. Wecan use limits to say this more precisely.



Definition of Infinite Limits


Definition (Infinite Limits)

Definition of Two Sided Infinite Limits






We say thatlimx→a

f (x) = ∞

if f (x) can be made arbitrarily large and positive by making xsufficiently close to a, on either side of a,







We say thatlimx→a

f (x) = ∞

if f (x) can be made arbitrarily large and positive by making xsufficiently close to a, on either side of a, and

limx→a

f (x) = −∞

if f (x) can be made arbitrarily large and negative by making xsufficiently close to a, on either side of a.







We say thatlimx→a

f (x) = ∞

if f (x) can be made arbitrarily large and positive by making xsufficiently close to a, on either side of a, and

limx→a

f (x) = −∞

if f (x) can be made arbitrarily large and negative by making xsufficiently close to a, on either side of a.

Similar definitions apply for one-sided infinite limits.




Example 5 – Evaluating Infinite Limits


For the function

f (x) =x + 1

x2 − 1

from Example 1 give the value as +∞ or −∞ for each limit, or explainwhy the infinite limit does not exist.

(a)

(b)

(c)





For the function

f (x) =x + 1

x2 − 1


(a) limx→1−

f (x)

(b) limx→1+

f (x)

(c) limx→1

f (x)





For the function

f (x) =x + 1

x2 − 1


(a) limx→1−

f (x) = −∞

(b) limx→1+

f (x)

(c) limx→1

f (x)





For the function

f (x) =x + 1

x2 − 1


(a) limx→1−

f (x) = −∞

(b) limx→1+

f (x) = ∞

(c) limx→1

f (x)





For the function

f (x) =x + 1

x2 − 1


(a) limx→1−

f (x) = −∞

(b) limx→1+

f (x) = ∞

(c) limx→1

f (x) DNE





For the function

f (x) =x + 1

x2 − 1


(a) limx→1−

f (x) = −∞

(b) limx→1+

f (x) = ∞

(c) limx→1

f (x) DNE

The two-sided limit at x = 1 does not exist (DNE) because the twoone-sided limits at x = 1 are different.


Appendix

More on Applying the Definition of Limit

In Example 2, if we wish to keep the value of f (x) within 0.03 of thelimit value, so that

−0.53 ≤ f (x) ≤ −0.47,

making the vertical extent of the box smaller, how much do we haveto decrease its horizontal extent?

Return


Appendix



−0.53 ≤ f (x) ≤ −0.47,

making the vertical extent of the box smaller, how much do we haveto decrease its horizontal extent?Using the values from the table in Example 1 we see that we onlyneed to have x within 0.1 of the limit point, i.e., −1.1 ≤ x ≤ −0.9.

Return


Appendix



−0.53 ≤ f (x) ≤ −0.47,

making the vertical extent of the box smaller, how much do we haveto decrease its horizontal extent?Using the values from the table in Example 1 we see that we onlyneed to have x within 0.1 of the limit point, i.e., −1.1 ≤ x ≤ −0.9.Since

f (−1.1) = −0.476190 < −0.47 and f (−0.9) = −0.526316 > −0.53

so that for x in the interval [−1.1,−0.9], the value of f is in theinterval [−0.526316,−0.476190], which is contained in the interval[−0.53,−0.47].

Return


Appendix



−0.53 ≤ f (x) ≤ −0.47,

making the vertical extent of the box smaller, how much do we haveto decrease its horizontal extent?Using the values from the table in Example 1 we see that we onlyneed to have x within 0.1 of the limit point, i.e., −1.1 ≤ x ≤ −0.9.Since

f (−1.1) = −0.476190 < −0.47 and f (−0.9) = −0.526316 > −0.53

so that for x in the interval [−1.1,−0.9], the value of f is in theinterval [−0.526316,−0.476190], which is contained in the interval[−0.53,−0.47].You draw the graph, by hand and on your calculator or using Maple,similar to the one on the previous slide.

Return


Appendix

Definition of One-Sided Infinite Limits

Definition (One-Sided Infinite Limits)

Return


Appendix



We say thatlim

x→a+f (x) = ∞

if f (x) can be made arbitrarily large and positive by making x suffi-ciently close to a with x > a,

Return


Appendix



We say thatlim

x→a+f (x) = ∞

if f (x) can be made arbitrarily large and positive by making x suffi-ciently close to a with x > a, and

limx→a−

f (x) = ∞

if f (x) can be made arbitrarily large and positive by making x suffi-ciently close to a with x < a.

Return


Appendix



We say thatlim

x→a+f (x) = −∞

if f (x) can be made arbitrarily large and negative by making x suffi-ciently close to a with x > a,

Return


Appendix



We say thatlim

x→a+f (x) = −∞

if f (x) can be made arbitrarily large and negative by making x suffi-ciently close to a with x > a, and

limx→a−

f (x) = −∞

if f (x) can be made arbitrarily large and negative by making x suffi-ciently close to a with x < a.

Return


Appendix



We say thatlim

x→a+f (x) = −∞

if f (x) can be made arbitrarily large and negative by making x suffi-ciently close to a with x > a, and

limx→a−

f (x) = −∞

if f (x) can be made arbitrarily large and negative by making x suffi-ciently close to a with x < a.

If the two one-sided infinite limits are different, then the two-sidedinfinite limit does not exist.

Return


Math 112 Lecture 1: Introduction to Limits - Okanagan...

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