lesson 4: calculating limits (section 21 slides)
DESCRIPTION
basic limits, limit laws, limits of algebraic and rational functions, limits of piecewise functions, limits of trigonometric quotients.TRANSCRIPT
Section 1.4Calculating Limits
V63.0121.021, Calculus I
New York University
September 16, 2010
Announcements
I First written homework due today (put it in the envelope) Rememberto put your lecture and recitation section numbers on your paper
Announcements
I First written homework duetoday (put it in theenvelope) Remember to putyour lecture and recitationsection numbers on yourpaper
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 2 / 45
Yoda on teaching a concepts course
“You must unlearn what you have learned.”
In other words, we are building up concepts and allowing ourselves only tospeak in terms of what we personally have produced.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 3 / 45
Objectives
I Know basic limits likelimx→a
x = a and limx→a
c = c .
I Use the limit laws tocompute elementary limits.
I Use algebra to simplifylimits.
I Understand and state theSqueeze Theorem.
I Use the Squeeze Theorem todemonstrate a limit.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 4 / 45
Outline
Recall: The concept of limit
Basic Limits
Limit LawsThe direct substitution property
Limits with AlgebraTwo more limit theorems
Two important trigonometric limits
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 5 / 45
Heuristic Definition of a Limit
Definition
We writelimx→a
f (x) = L
and say
“the limit of f (x), as x approaches a, equals L”
if we can make the values of f (x) arbitrarily close to L (as close to L as welike) by taking x to be sufficiently close to a (on either side of a) but notequal to a.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 6 / 45
The error-tolerance game
A game between two players (Dana and Emerson) to decide if a limitlimx→a
f (x) exists.
Step 1 Dana proposes L to be the limit.
Step 2 Emerson challenges with an “error” level around L.
Step 3 Dana chooses a “tolerance” level around a so that points x withinthat tolerance of a (not counting a itself) are taken to values ywithin the error level of L. If Dana cannot, Emerson wins and thelimit cannot be L.
Step 4 If Dana’s move is a good one, Emerson can challenge again or giveup. If Emerson gives up, Dana wins and the limit is L.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 7 / 45
The error-tolerance game
This tolerance is too bigStill too bigThis looks goodSo does this
a
L
I To be legit, the part of the graph inside the blue (vertical) strip mustalso be inside the green (horizontal) strip.
I If Emerson shrinks the error, Dana can still win.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 8 / 45
Limit FAIL: Jump
x
y
−1
1
Part of graph in-side blue is notinside green
Part of graph in-side blue is notinside green
I So limx→0
|x |x
does not exist.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 9 / 45
Limit FAIL: Jump
x
y
−1
1
Part of graph in-side blue is notinside green
Part of graph in-side blue is notinside green
I So limx→0
|x |x
does not exist.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 9 / 45
Limit FAIL: Jump
x
y
−1
1
Part of graph in-side blue is notinside green
Part of graph in-side blue is notinside green
I So limx→0
|x |x
does not exist.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 9 / 45
Limit FAIL: unboundedness
x
y
0
L?
The graph escapes thegreen, so no good
Even worse!
limx→0+
1
xdoes not exist
because the function isunbounded near 0
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 10 / 45
Limit EPIC FAIL
Here is a graph of the function f (x) = sin(πx
):
x
y
−1
1
For every y in [−1, 1], there are infinitely many points x arbitrarily close tozero where f (x) = y . So lim
x→0f (x) cannot exist.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 11 / 45
Outline
Recall: The concept of limit
Basic Limits
Limit LawsThe direct substitution property
Limits with AlgebraTwo more limit theorems
Two important trigonometric limits
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 12 / 45
Really basic limits
Fact
Let c be a constant and a a real number.
(i) limx→a
x = a
(ii) limx→a
c = c
Proof.
The first is tautological, the second is trivial.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 13 / 45
Really basic limits
Fact
Let c be a constant and a a real number.
(i) limx→a
x = a
(ii) limx→a
c = c
Proof.
The first is tautological, the second is trivial.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 13 / 45
ET game for f (x) = x
x
y
a
a
I Setting error equal to tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 14 / 45
ET game for f (x) = x
x
y
a
a
I Setting error equal to tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 14 / 45
ET game for f (x) = x
x
y
a
a
I Setting error equal to tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 14 / 45
ET game for f (x) = x
x
y
a
a
I Setting error equal to tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 14 / 45
ET game for f (x) = x
x
y
a
a
I Setting error equal to tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 14 / 45
ET game for f (x) = x
x
y
a
a
I Setting error equal to tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 14 / 45
ET game for f (x) = x
x
y
a
a
I Setting error equal to tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 14 / 45
ET game for f (x) = c
x
y
a
c
I any tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 15 / 45
ET game for f (x) = c
x
y
a
c
I any tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 15 / 45
ET game for f (x) = c
x
y
a
c
I any tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 15 / 45
ET game for f (x) = c
x
y
a
c
I any tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 15 / 45
ET game for f (x) = c
x
y
a
c
I any tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 15 / 45
ET game for f (x) = c
x
y
a
c
I any tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 15 / 45
ET game for f (x) = c
x
y
a
c
I any tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 15 / 45
Really basic limits
Fact
Let c be a constant and a a real number.
(i) limx→a
x = a
(ii) limx→a
c = c
Proof.
The first is tautological, the second is trivial.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 16 / 45
Outline
Recall: The concept of limit
Basic Limits
Limit LawsThe direct substitution property
Limits with AlgebraTwo more limit theorems
Two important trigonometric limits
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 17 / 45
Limits and arithmetic
Fact
Suppose limx→a
f (x) = L and limx→a
g(x) = M and c is a constant. Then
1. limx→a
[f (x) + g(x)] = L + M
(errors add)
2. limx→a
[f (x)− g(x)] = L−M
(combination of adding and scaling)
3. limx→a
[cf (x)] = cL
(error scales)
4. limx→a
[f (x)g(x)] = L ·M
(more complicated, but doable)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 18 / 45
Limits and arithmetic
Fact
Suppose limx→a
f (x) = L and limx→a
g(x) = M and c is a constant. Then
1. limx→a
[f (x) + g(x)] = L + M (errors add)
2. limx→a
[f (x)− g(x)] = L−M
(combination of adding and scaling)
3. limx→a
[cf (x)] = cL
(error scales)
4. limx→a
[f (x)g(x)] = L ·M
(more complicated, but doable)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 18 / 45
Limits and arithmetic
Fact
Suppose limx→a
f (x) = L and limx→a
g(x) = M and c is a constant. Then
1. limx→a
[f (x) + g(x)] = L + M (errors add)
2. limx→a
[f (x)− g(x)] = L−M
(combination of adding and scaling)
3. limx→a
[cf (x)] = cL
(error scales)
4. limx→a
[f (x)g(x)] = L ·M
(more complicated, but doable)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 19 / 45
Limits and arithmetic
Fact
Suppose limx→a
f (x) = L and limx→a
g(x) = M and c is a constant. Then
1. limx→a
[f (x) + g(x)] = L + M (errors add)
2. limx→a
[f (x)− g(x)] = L−M
(combination of adding and scaling)
3. limx→a
[cf (x)] = cL
(error scales)
4. limx→a
[f (x)g(x)] = L ·M
(more complicated, but doable)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 19 / 45
Limits and arithmetic
Fact
Suppose limx→a
f (x) = L and limx→a
g(x) = M and c is a constant. Then
1. limx→a
[f (x) + g(x)] = L + M (errors add)
2. limx→a
[f (x)− g(x)] = L−M
(combination of adding and scaling)
3. limx→a
[cf (x)] = cL (error scales)
4. limx→a
[f (x)g(x)] = L ·M
(more complicated, but doable)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 19 / 45
Justification of the scaling law
I errors scale: If f (x) is e away from L, then
(c · f (x)− c · L) = c · (f (x)− L) = c · e
That is, (c · f )(x) is c · e away from cL,
I So if Emerson gives us an error of 1 (for instance), Dana can use thefact that lim
x→af (x) = L to find a tolerance for f and g corresponding
to the error 1/c.
I Dana wins the round.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 20 / 45
Limits and arithmetic
Fact
Suppose limx→a
f (x) = L and limx→a
g(x) = M and c is a constant. Then
1. limx→a
[f (x) + g(x)] = L + M (errors add)
2. limx→a
[f (x)− g(x)] = L−M (combination of adding and scaling)
3. limx→a
[cf (x)] = cL (error scales)
4. limx→a
[f (x)g(x)] = L ·M
(more complicated, but doable)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 21 / 45
Limits and arithmetic
Fact
Suppose limx→a
f (x) = L and limx→a
g(x) = M and c is a constant. Then
1. limx→a
[f (x) + g(x)] = L + M (errors add)
2. limx→a
[f (x)− g(x)] = L−M (combination of adding and scaling)
3. limx→a
[cf (x)] = cL (error scales)
4. limx→a
[f (x)g(x)] = L ·M
(more complicated, but doable)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 22 / 45
Limits and arithmetic
Fact
Suppose limx→a
f (x) = L and limx→a
g(x) = M and c is a constant. Then
1. limx→a
[f (x) + g(x)] = L + M (errors add)
2. limx→a
[f (x)− g(x)] = L−M (combination of adding and scaling)
3. limx→a
[cf (x)] = cL (error scales)
4. limx→a
[f (x)g(x)] = L ·M (more complicated, but doable)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 22 / 45
Limits and arithmetic II
Fact (Continued)
5. limx→a
f (x)
g(x)=
L
M, if M 6= 0.
6. limx→a
[f (x)]n =[
limx→a
f (x)]n
(follows from 4 repeatedly)
7. limx→a
xn = an
(follows from 6)
8. limx→a
n√x = n√a
9. limx→a
n√f (x) = n
√limx→a
f (x) (If n is even, we must additionally assume
that limx→a
f (x) > 0)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 23 / 45
Caution!
I The quotient rule for limits says that if limx→a
g(x) 6= 0, then
limx→a
f (x)
g(x)=
limx→a f (x)
limx→a g(x)
I It does NOT say that if limx→a
g(x) = 0, then
limx→a
f (x)
g(x)does not exist
In fact, limits of quotients where numerator and denominator bothtend to 0 are exactly where the magic happens.
I more about this later
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 24 / 45
Limits and arithmetic II
Fact (Continued)
5. limx→a
f (x)
g(x)=
L
M, if M 6= 0.
6. limx→a
[f (x)]n =[
limx→a
f (x)]n
(follows from 4 repeatedly)
7. limx→a
xn = an
(follows from 6)
8. limx→a
n√x = n√a
9. limx→a
n√f (x) = n
√limx→a
f (x) (If n is even, we must additionally assume
that limx→a
f (x) > 0)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
Limits and arithmetic II
Fact (Continued)
5. limx→a
f (x)
g(x)=
L
M, if M 6= 0.
6. limx→a
[f (x)]n =[
limx→a
f (x)]n
(follows from 4 repeatedly)
7. limx→a
xn = an
(follows from 6)
8. limx→a
n√x = n√a
9. limx→a
n√f (x) = n
√limx→a
f (x) (If n is even, we must additionally assume
that limx→a
f (x) > 0)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
Limits and arithmetic II
Fact (Continued)
5. limx→a
f (x)
g(x)=
L
M, if M 6= 0.
6. limx→a
[f (x)]n =[
limx→a
f (x)]n
(follows from 4 repeatedly)
7. limx→a
xn = an
(follows from 6)
8. limx→a
n√x = n√a
9. limx→a
n√f (x) = n
√limx→a
f (x) (If n is even, we must additionally assume
that limx→a
f (x) > 0)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
Limits and arithmetic II
Fact (Continued)
5. limx→a
f (x)
g(x)=
L
M, if M 6= 0.
6. limx→a
[f (x)]n =[
limx→a
f (x)]n
(follows from 4 repeatedly)
7. limx→a
xn = an
(follows from 6)
8. limx→a
n√x = n√a
9. limx→a
n√f (x) = n
√limx→a
f (x) (If n is even, we must additionally assume
that limx→a
f (x) > 0)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
Limits and arithmetic II
Fact (Continued)
5. limx→a
f (x)
g(x)=
L
M, if M 6= 0.
6. limx→a
[f (x)]n =[
limx→a
f (x)]n
(follows from 4 repeatedly)
7. limx→a
xn = an (follows from 6)
8. limx→a
n√x = n√a
9. limx→a
n√f (x) = n
√limx→a
f (x) (If n is even, we must additionally assume
that limx→a
f (x) > 0)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
Limits and arithmetic II
Fact (Continued)
5. limx→a
f (x)
g(x)=
L
M, if M 6= 0.
6. limx→a
[f (x)]n =[
limx→a
f (x)]n
(follows from 4 repeatedly)
7. limx→a
xn = an (follows from 6)
8. limx→a
n√x = n√a
9. limx→a
n√
f (x) = n
√limx→a
f (x) (If n is even, we must additionally assume
that limx→a
f (x) > 0)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
Applying the limit laws
Example
Find limx→3
(x2 + 2x + 4
).
Solution
By applying the limit laws repeatedly:
limx→3
(x2 + 2x + 4
)
= limx→3
(x2)
+ limx→3
(2x) + limx→3
(4)
=(
limx→3
x)2
+ 2 · limx→3
(x) + 4
= (3)2 + 2 · 3 + 4
= 9 + 6 + 4 = 19.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
Applying the limit laws
Example
Find limx→3
(x2 + 2x + 4
).
Solution
By applying the limit laws repeatedly:
limx→3
(x2 + 2x + 4
)
= limx→3
(x2)
+ limx→3
(2x) + limx→3
(4)
=(
limx→3
x)2
+ 2 · limx→3
(x) + 4
= (3)2 + 2 · 3 + 4
= 9 + 6 + 4 = 19.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
Applying the limit laws
Example
Find limx→3
(x2 + 2x + 4
).
Solution
By applying the limit laws repeatedly:
limx→3
(x2 + 2x + 4
)= lim
x→3
(x2)
+ limx→3
(2x) + limx→3
(4)
=(
limx→3
x)2
+ 2 · limx→3
(x) + 4
= (3)2 + 2 · 3 + 4
= 9 + 6 + 4 = 19.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
Applying the limit laws
Example
Find limx→3
(x2 + 2x + 4
).
Solution
By applying the limit laws repeatedly:
limx→3
(x2 + 2x + 4
)= lim
x→3
(x2)
+ limx→3
(2x) + limx→3
(4)
=(
limx→3
x)2
+ 2 · limx→3
(x) + 4
= (3)2 + 2 · 3 + 4
= 9 + 6 + 4 = 19.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
Applying the limit laws
Example
Find limx→3
(x2 + 2x + 4
).
Solution
By applying the limit laws repeatedly:
limx→3
(x2 + 2x + 4
)= lim
x→3
(x2)
+ limx→3
(2x) + limx→3
(4)
=(
limx→3
x)2
+ 2 · limx→3
(x) + 4
= (3)2 + 2 · 3 + 4
= 9 + 6 + 4 = 19.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
Applying the limit laws
Example
Find limx→3
(x2 + 2x + 4
).
Solution
By applying the limit laws repeatedly:
limx→3
(x2 + 2x + 4
)= lim
x→3
(x2)
+ limx→3
(2x) + limx→3
(4)
=(
limx→3
x)2
+ 2 · limx→3
(x) + 4
= (3)2 + 2 · 3 + 4
= 9 + 6 + 4 = 19.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
Your turn
Example
Find limx→3
x2 + 2x + 4
x3 + 11
Solution
The answer is19
38=
1
2.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 27 / 45
Your turn
Example
Find limx→3
x2 + 2x + 4
x3 + 11
Solution
The answer is19
38=
1
2.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 27 / 45
Direct Substitution Property
Theorem (The Direct Substitution Property)
If f is a polynomial or a rational function and a is in the domain of f , then
limx→a
f (x) = f (a)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 28 / 45
Outline
Recall: The concept of limit
Basic Limits
Limit LawsThe direct substitution property
Limits with AlgebraTwo more limit theorems
Two important trigonometric limits
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 29 / 45
Limits do not see the point! (in a good way)
Theorem
If f (x) = g(x) when x 6= a, and limx→a
g(x) = L, then limx→a
f (x) = L.
Example
Find limx→−1
x2 + 2x + 1
x + 1, if it exists.
Solution
Sincex2 + 2x + 1
x + 1= x + 1 whenever x 6= −1, and since lim
x→−1x + 1 = 0,
we have limx→−1
x2 + 2x + 1
x + 1= 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 30 / 45
Limits do not see the point! (in a good way)
Theorem
If f (x) = g(x) when x 6= a, and limx→a
g(x) = L, then limx→a
f (x) = L.
Example
Find limx→−1
x2 + 2x + 1
x + 1, if it exists.
Solution
Sincex2 + 2x + 1
x + 1= x + 1 whenever x 6= −1, and since lim
x→−1x + 1 = 0,
we have limx→−1
x2 + 2x + 1
x + 1= 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 30 / 45
Limits do not see the point! (in a good way)
Theorem
If f (x) = g(x) when x 6= a, and limx→a
g(x) = L, then limx→a
f (x) = L.
Example
Find limx→−1
x2 + 2x + 1
x + 1, if it exists.
Solution
Sincex2 + 2x + 1
x + 1= x + 1 whenever x 6= −1, and since lim
x→−1x + 1 = 0,
we have limx→−1
x2 + 2x + 1
x + 1= 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 30 / 45
ET game for f (x) =x2 + 2x + 1
x + 1
x
y
−1
I Even if f (−1) were something else, it would not effect the limit.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 31 / 45
ET game for f (x) =x2 + 2x + 1
x + 1
x
y
−1
I Even if f (−1) were something else, it would not effect the limit.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 31 / 45
Limit of a function defined piecewise at a boundarypoint
Example
Let
f (x) =
{x2 x ≥ 0
−x x < 0
Does limx→0
f (x) exist?
Solution
We havelim
x→0+f (x)
MTP= lim
x→0+x2
DSP= 02 = 0
Likewise:lim
x→0−f (x) = lim
x→0−−x = −0 = 0
So limx→0
f (x) = 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
Limit of a function defined piecewise at a boundarypoint
Example
Let
f (x) =
{x2 x ≥ 0
−x x < 0
Does limx→0
f (x) exist?
Solution
We havelim
x→0+f (x)
MTP= lim
x→0+x2
DSP= 02 = 0
Likewise:lim
x→0−f (x) = lim
x→0−−x = −0 = 0
So limx→0
f (x) = 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
Limit of a function defined piecewise at a boundarypoint
Example
Let
f (x) =
{x2 x ≥ 0
−x x < 0
Does limx→0
f (x) exist?
Solution
We havelim
x→0+f (x)
MTP= lim
x→0+x2
DSP= 02 = 0
Likewise:lim
x→0−f (x) = lim
x→0−−x = −0 = 0
So limx→0
f (x) = 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
Limit of a function defined piecewise at a boundarypoint
Example
Let
f (x) =
{x2 x ≥ 0
−x x < 0
Does limx→0
f (x) exist?
Solution
We havelim
x→0+f (x)
MTP= lim
x→0+x2
DSP= 02 = 0
Likewise:lim
x→0−f (x) = lim
x→0−−x = −0 = 0
So limx→0
f (x) = 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
Limit of a function defined piecewise at a boundarypoint
Example
Let
f (x) =
{x2 x ≥ 0
−x x < 0
Does limx→0
f (x) exist?
Solution
We havelim
x→0+f (x)
MTP= lim
x→0+x2
DSP= 02 = 0
Likewise:lim
x→0−f (x) = lim
x→0−−x = −0 = 0
So limx→0
f (x) = 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
Limit of a function defined piecewise at a boundarypoint
Example
Let
f (x) =
{x2 x ≥ 0
−x x < 0
Does limx→0
f (x) exist?
Solution
We havelim
x→0+f (x)
MTP= lim
x→0+x2
DSP= 02 = 0
Likewise:lim
x→0−f (x) = lim
x→0−−x = −0 = 0
So limx→0
f (x) = 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
Limit of a function defined piecewise at a boundarypoint
Example
Let
f (x) =
{x2 x ≥ 0
−x x < 0
Does limx→0
f (x) exist?
Solution
We havelim
x→0+f (x)
MTP= lim
x→0+x2
DSP= 02 = 0
Likewise:lim
x→0−f (x) = lim
x→0−−x = −0 = 0
So limx→0
f (x) = 0.V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
Finding limits by algebraic manipulations
Example
Find limx→4
√x − 2
x − 4.
Solution
Write the denominator as x − 4 =√x2 − 4 = (
√x − 2)(
√x + 2). So
limx→4
√x − 2
x − 4= lim
x→4
√x − 2
(√x − 2)(
√x + 2)
= limx→4
1√x + 2
=1
4
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 33 / 45
Finding limits by algebraic manipulations
Example
Find limx→4
√x − 2
x − 4.
Solution
Write the denominator as x − 4 =√x2 − 4 = (
√x − 2)(
√x + 2).
So
limx→4
√x − 2
x − 4= lim
x→4
√x − 2
(√x − 2)(
√x + 2)
= limx→4
1√x + 2
=1
4
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 33 / 45
Finding limits by algebraic manipulations
Example
Find limx→4
√x − 2
x − 4.
Solution
Write the denominator as x − 4 =√x2 − 4 = (
√x − 2)(
√x + 2). So
limx→4
√x − 2
x − 4= lim
x→4
√x − 2
(√x − 2)(
√x + 2)
= limx→4
1√x + 2
=1
4
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 33 / 45
Your turn
Example
Let
f (x) =
{1− x2 x ≥ 1
2x x < 1
Find limx→1
f (x) if it exists.
1
Solution
We have
limx→1+
f (x) = limx→1+
(1− x2
) DSP= 0
limx→1−
f (x) = limx→1−
(2x)DSP= 2
The left- and right-hand limits disagree, so the limit does not exist.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
Your turn
Example
Let
f (x) =
{1− x2 x ≥ 1
2x x < 1
Find limx→1
f (x) if it exists.
1
Solution
We have
limx→1+
f (x) = limx→1+
(1− x2
) DSP= 0
limx→1−
f (x) = limx→1−
(2x)DSP= 2
The left- and right-hand limits disagree, so the limit does not exist.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
Your turn
Example
Let
f (x) =
{1− x2 x ≥ 1
2x x < 1
Find limx→1
f (x) if it exists. 1
Solution
We have
limx→1+
f (x) = limx→1+
(1− x2
) DSP= 0
limx→1−
f (x) = limx→1−
(2x)DSP= 2
The left- and right-hand limits disagree, so the limit does not exist.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
Your turn
Example
Let
f (x) =
{1− x2 x ≥ 1
2x x < 1
Find limx→1
f (x) if it exists. 1
Solution
We have
limx→1+
f (x) = limx→1+
(1− x2
) DSP= 0
limx→1−
f (x) = limx→1−
(2x)DSP= 2
The left- and right-hand limits disagree, so the limit does not exist.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
Your turn
Example
Let
f (x) =
{1− x2 x ≥ 1
2x x < 1
Find limx→1
f (x) if it exists. 1
Solution
We have
limx→1+
f (x) = limx→1+
(1− x2
) DSP= 0
limx→1−
f (x) = limx→1−
(2x)DSP= 2
The left- and right-hand limits disagree, so the limit does not exist.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
Your turn
Example
Let
f (x) =
{1− x2 x ≥ 1
2x x < 1
Find limx→1
f (x) if it exists. 1
Solution
We have
limx→1+
f (x) = limx→1+
(1− x2
) DSP= 0
limx→1−
f (x) = limx→1−
(2x)DSP= 2
The left- and right-hand limits disagree, so the limit does not exist.V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
A message from the Mathematical Grammar Police
Please do not say “ limx→a
f (x) = DNE.” Does not compute.
I Too many verbs
I Leads to FALSE limit laws like “If limx→a
f (x) DNE and limx→a
g(x) DNE,
then limx→a
(f (x) + g(x)) DNE.”
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 35 / 45
A message from the Mathematical Grammar Police
Please do not say “ limx→a
f (x) = DNE.” Does not compute.
I Too many verbs
I Leads to FALSE limit laws like “If limx→a
f (x) DNE and limx→a
g(x) DNE,
then limx→a
(f (x) + g(x)) DNE.”
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 35 / 45
A message from the Mathematical Grammar Police
Please do not say “ limx→a
f (x) = DNE.” Does not compute.
I Too many verbs
I Leads to FALSE limit laws like “If limx→a
f (x) DNE and limx→a
g(x) DNE,
then limx→a
(f (x) + g(x)) DNE.”
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 35 / 45
Two More Important Limit Theorems
Theorem
If f (x) ≤ g(x) when x is near a (except possibly at a), then
limx→a
f (x) ≤ limx→a
g(x)
(as usual, provided these limits exist).
Theorem (The Squeeze/Sandwich/Pinching Theorem)
If f (x) ≤ g(x) ≤ h(x) when x is near a (as usual, except possibly at a),and
limx→a
f (x) = limx→a
h(x) = L,
thenlimx→a
g(x) = L.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 36 / 45
Using the Squeeze Theorem
We can use the Squeeze Theorem to replace complicated expressions withsimple ones when taking the limit.
Example
Show that limx→0
x2 sin(πx
)= 0.
Solution
We have for all x,
−1 ≤ sin(πx
)≤ 1 =⇒ −x2 ≤ x2 sin
(πx
)≤ x2
The left and right sides go to zero as x → 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 37 / 45
Using the Squeeze Theorem
We can use the Squeeze Theorem to replace complicated expressions withsimple ones when taking the limit.
Example
Show that limx→0
x2 sin(πx
)= 0.
Solution
We have for all x,
−1 ≤ sin(πx
)≤ 1 =⇒ −x2 ≤ x2 sin
(πx
)≤ x2
The left and right sides go to zero as x → 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 37 / 45
Using the Squeeze Theorem
We can use the Squeeze Theorem to replace complicated expressions withsimple ones when taking the limit.
Example
Show that limx→0
x2 sin(πx
)= 0.
Solution
We have for all x,
−1 ≤ sin(πx
)≤ 1 =⇒ −x2 ≤ x2 sin
(πx
)≤ x2
The left and right sides go to zero as x → 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 37 / 45
Illustration of the Squeeze Theorem
x
y h(x) = x2
f (x) = −x2
g(x) = x2 sin(πx
)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 38 / 45
Illustration of the Squeeze Theorem
x
y h(x) = x2
f (x) = −x2
g(x) = x2 sin(πx
)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 38 / 45
Illustration of the Squeeze Theorem
x
y h(x) = x2
f (x) = −x2
g(x) = x2 sin(πx
)
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 38 / 45
Outline
Recall: The concept of limit
Basic Limits
Limit LawsThe direct substitution property
Limits with AlgebraTwo more limit theorems
Two important trigonometric limits
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 39 / 45
Two important trigonometric limits
Theorem
The following two limits hold:
I limθ→0
sin θ
θ= 1
I limθ→0
cos θ − 1
θ= 0
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 40 / 45
Proof of the Sine Limit
Proof.
θ
sin θ
cos θ
θ
tan θ
−1 1
Notice
sin θ ≤
θ
≤ 2 tanθ
2≤ tan θ
Divide by sin θ:
1 ≤ θ
sin θ≤ 1
cos θ
Take reciprocals:
1 ≥ sin θ
θ≥ cos θ
As θ → 0, the left and right sides tend to 1. So, then, must the middleexpression.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
Proof of the Sine Limit
Proof.
θsin θ
cos θ
θ
tan θ
−1 1
Notice
sin θ ≤ θ
≤ 2 tanθ
2≤ tan θ
Divide by sin θ:
1 ≤ θ
sin θ≤ 1
cos θ
Take reciprocals:
1 ≥ sin θ
θ≥ cos θ
As θ → 0, the left and right sides tend to 1. So, then, must the middleexpression.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
Proof of the Sine Limit
Proof.
θsin θ
cos θ
θ tan θ
−1 1
Notice
sin θ ≤ θ
≤ 2 tanθ
2≤
tan θ
Divide by sin θ:
1 ≤ θ
sin θ≤ 1
cos θ
Take reciprocals:
1 ≥ sin θ
θ≥ cos θ
As θ → 0, the left and right sides tend to 1. So, then, must the middleexpression.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
Proof of the Sine Limit
Proof.
θsin θ
cos θ
θ tan θ
−1 1
Notice
sin θ ≤ θ ≤ 2 tanθ
2≤ tan θ
Divide by sin θ:
1 ≤ θ
sin θ≤ 1
cos θ
Take reciprocals:
1 ≥ sin θ
θ≥ cos θ
As θ → 0, the left and right sides tend to 1. So, then, must the middleexpression.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
Proof of the Sine Limit
Proof.
θsin θ
cos θ
θ tan θ
−1 1
Notice
sin θ ≤ θ ≤ 2 tanθ
2≤ tan θ
Divide by sin θ:
1 ≤ θ
sin θ≤ 1
cos θ
Take reciprocals:
1 ≥ sin θ
θ≥ cos θ
As θ → 0, the left and right sides tend to 1. So, then, must the middleexpression.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
Proof of the Sine Limit
Proof.
θsin θ
cos θ
θ tan θ
−1 1
Notice
sin θ ≤ θ ≤ 2 tanθ
2≤ tan θ
Divide by sin θ:
1 ≤ θ
sin θ≤ 1
cos θ
Take reciprocals:
1 ≥ sin θ
θ≥ cos θ
As θ → 0, the left and right sides tend to 1. So, then, must the middleexpression.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
Proof of the Sine Limit
Proof.
θsin θ
cos θ
θ tan θ
−1 1
Notice
sin θ ≤ θ ≤ 2 tanθ
2≤ tan θ
Divide by sin θ:
1 ≤ θ
sin θ≤ 1
cos θ
Take reciprocals:
1 ≥ sin θ
θ≥ cos θ
As θ → 0, the left and right sides tend to 1. So, then, must the middleexpression.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
Proof of the Cosine Limit
Proof.
1− cos θ
θ=
1− cos θ
θ· 1 + cos θ
1 + cos θ=
1− cos2 θ
θ(1 + cos θ)
=sin2 θ
θ(1 + cos θ)=
sin θ
θ· sin θ
1 + cos θ
So
limθ→0
1− cos θ
θ=
(limθ→0
sin θ
θ
)·(
limθ→0
sin θ
1 + cos θ
)= 1 · 0 = 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 42 / 45
Try these
Example
1. limθ→0
tan θ
θ
2. limθ→0
sin 2θ
θ
Answer
1. 1
2. 2
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 43 / 45
Try these
Example
1. limθ→0
tan θ
θ
2. limθ→0
sin 2θ
θ
Answer
1. 1
2. 2
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 43 / 45
Solutions
1. Use the basic trigonometric limit and the definition of tangent.
limθ→0
tan θ
θ= lim
θ→0
sin θ
θ cos θ= lim
θ→0
sin θ
θ· limθ→0
1
cos θ= 1 · 1
1= 1.
2. Change the variable:
limθ→0
sin 2θ
θ= lim
2θ→0
sin 2θ
2θ · 12= 2 · lim
2θ→0
sin 2θ
2θ= 2 · 1 = 2
OR use a trigonometric identity:
limθ→0
sin 2θ
θ= lim
θ→0
2 sin θ cos θ
θ= 2 · lim
θ→0
sin θ
θ· limθ→0
cos θ = 2 · 1 · 1 = 2
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 44 / 45
Solutions
1. Use the basic trigonometric limit and the definition of tangent.
limθ→0
tan θ
θ= lim
θ→0
sin θ
θ cos θ= lim
θ→0
sin θ
θ· limθ→0
1
cos θ= 1 · 1
1= 1.
2. Change the variable:
limθ→0
sin 2θ
θ= lim
2θ→0
sin 2θ
2θ · 12= 2 · lim
2θ→0
sin 2θ
2θ= 2 · 1 = 2
OR use a trigonometric identity:
limθ→0
sin 2θ
θ= lim
θ→0
2 sin θ cos θ
θ= 2 · lim
θ→0
sin θ
θ· limθ→0
cos θ = 2 · 1 · 1 = 2
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 44 / 45
Solutions
1. Use the basic trigonometric limit and the definition of tangent.
limθ→0
tan θ
θ= lim
θ→0
sin θ
θ cos θ= lim
θ→0
sin θ
θ· limθ→0
1
cos θ= 1 · 1
1= 1.
2. Change the variable:
limθ→0
sin 2θ
θ= lim
2θ→0
sin 2θ
2θ · 12= 2 · lim
2θ→0
sin 2θ
2θ= 2 · 1 = 2
OR use a trigonometric identity:
limθ→0
sin 2θ
θ= lim
θ→0
2 sin θ cos θ
θ= 2 · lim
θ→0
sin θ
θ· limθ→0
cos θ = 2 · 1 · 1 = 2
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 44 / 45
Summary
I The limit laws allow us to compute limits reasonably.
I BUT we cannot make up extra laws otherwise we get into trouble.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 45 / 45