Lesson 4 - Calculating Limits (Slides+Notes)
32
Section 2.3 Computation of Limits Math 1a October 1, 2007
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Transcript of Lesson 4 - Calculating Limits (Slides+Notes)
Section 2.3Computation of Limits
Math 1a
October 1, 2007
Math 1a - October 01, 2007.GWBMonday, Oct 1, 2007
Page1of13
Limit Laws
Suppose that c is a constant and the limits
limx→a
f (x) and limx→a
g(x)
exist. Then
1. 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)
Math 1a - October 01, 2007.GWBMonday, Oct 1, 2007
Page2of13
Limit Laws, continued
5. limx→a
f (x)
g(x)=
limx→a
f (x)
limx→a
g(x), if lim
x→ag(x) 6= 0.
6. limx→a
[f (x)]n =[limx→a
f (x)]n
(follows from 3 repeatedly)
7. limx→a
c = c
8. limx→a
x = a
9. limx→a
xn = an
(follows from 6 and 8)
10. limx→a
n√
x = n√
a
11. limx→a
n√
f (x) = n
√limx→a
f (x) (If n is even, we must additionally
assume that limx→a
f (x) > 0)
Limit Laws, continued
5. limx→a
f (x)
g(x)=
limx→a
f (x)
limx→a
g(x), if lim
x→ag(x) 6= 0.
6. limx→a
[f (x)]n =[limx→a
f (x)]n
(follows from 3 repeatedly)
7. limx→a
c = c
8. limx→a
x = a
9. limx→a
xn = an
(follows from 6 and 8)
10. limx→a
n√
x = n√
a
11. limx→a
n√
f (x) = n
√limx→a
f (x) (If n is even, we must additionally
assume that limx→a
f (x) > 0)
Math 1a - October 01, 2007.GWBMonday, Oct 1, 2007
Page3of13
Limit Laws, continued
5. limx→a
f (x)
g(x)=
limx→a
f (x)
limx→a
g(x), if lim
x→ag(x) 6= 0.
6. limx→a
[f (x)]n =[limx→a
f (x)]n
(follows from 3 repeatedly)
7. limx→a
c = c
8. limx→a
x = a
9. limx→a
xn = an
(follows from 6 and 8)
10. limx→a
n√
x = n√
a
11. limx→a
n√
f (x) = n
√limx→a
f (x) (If n is even, we must additionally
assume that limx→a
f (x) > 0)
Limit Laws, continued
5. limx→a
f (x)
g(x)=
limx→a
f (x)
limx→a
g(x), if lim
x→ag(x) 6= 0.
6. limx→a
[f (x)]n =[limx→a
f (x)]n
(follows from 3 repeatedly)
7. limx→a
c = c
8. limx→a
x = a
9. limx→a
xn = an
(follows from 6 and 8)
10. limx→a
n√
x = n√
a
11. limx→a
n√
f (x) = n
√limx→a
f (x) (If n is even, we must additionally
assume that limx→a
f (x) > 0)
Math 1a - October 01, 2007.GWBMonday, Oct 1, 2007
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Math 1a - October 01, 2007.GWBMonday, Oct 1, 2007
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Math 1a - October 01, 2007.GWBMonday, Oct 1, 2007
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Math 1a - October 01, 2007.GWBMonday, Oct 1, 2007
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Math 1a - October 01, 2007.GWBMonday, Oct 1, 2007
Page8of13
Limit Laws, continued
5. limx→a
f (x)
g(x)=
limx→a
f (x)
limx→a
g(x), if lim
x→ag(x) 6= 0.
6. limx→a
[f (x)]n =[limx→a
f (x)]n
(follows from 3 repeatedly)
7. limx→a
c = c
8. limx→a
x = a
9. limx→a
xn = an
(follows from 6 and 8)
10. limx→a
n√
x = n√
a
11. limx→a
n√
f (x) = n
√limx→a
f (x) (If n is even, we must additionally
assume that limx→a
f (x) > 0)
Limit Laws, continued
5. limx→a
f (x)
g(x)=
limx→a
f (x)
limx→a
g(x), if lim
x→ag(x) 6= 0.
6. limx→a
[f (x)]n =[limx→a
f (x)]n
(follows from 3 repeatedly)
7. limx→a
c = c
8. limx→a
x = a
9. limx→a
xn = an (follows from 6 and 8)
10. limx→a
n√
x = n√
a
11. limx→a
n√
f (x) = n
√limx→a
f (x) (If n is even, we must additionally
assume that limx→a
f (x) > 0)
Limit Laws, continued
5. limx→a
f (x)
g(x)=
limx→a
f (x)
limx→a
g(x), if lim
x→ag(x) 6= 0.
6. limx→a
[f (x)]n =[limx→a
f (x)]n
(follows from 3 repeatedly)
7. limx→a
c = c
8. limx→a
x = a
9. limx→a
xn = an (follows from 6 and 8)
10. limx→a
n√
x = n√
a
11. limx→a
n√
f (x) = n
√limx→a
f (x) (If n is even, we must additionally
assume that limx→a
f (x) > 0)
Direct Substitution Property
Theorem (The Direct Substitution Property)
If f is a polynomial or a rational function and a is in the domain off , then
limx→a
f (x) = f (a)
Limits do not see the point! (in a good way)
TheoremIf f (x) = g(x) when x 6= a, and lim
x→ag(x) = L, then lim
x→af (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
limx→−1
x + 1 = 0, we have limx→−1
x2 + 2x + 1
x + 1= 0.
Math 1a - October 01, 2007.GWBMonday, Oct 1, 2007
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Limits do not see the point! (in a good way)
TheoremIf f (x) = g(x) when x 6= a, and lim
x→ag(x) = L, then lim
x→af (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
limx→−1
x + 1 = 0, we have limx→−1
x2 + 2x + 1
x + 1= 0.
Limits do not see the point! (in a good way)
TheoremIf f (x) = g(x) when x 6= a, and lim
x→ag(x) = L, then lim
x→af (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
limx→−1
x + 1 = 0, we have limx→−1
x2 + 2x + 1
x + 1= 0.
Finding limits by algebraic manipulations
Example
Find limx→4
√x − 2
x − 4.
SolutionWrite the denominator as x − 4 =
√x
2 − 4 = (√
x − 2)(√
x + 2).So
limx→2
√x − 2
x − 4= lim
x→2
√x − 2
(√
x − 2)(√
x + 2)
= limx→2
1√x + 2
=1
4
Example
Try limx→a
3√
x − 3√
a
x − a.
Finding limits by algebraic manipulations
Example
Find limx→4
√x − 2
x − 4.
SolutionWrite the denominator as x − 4 =
√x
2 − 4 = (√
x − 2)(√
x + 2).
So
limx→2
√x − 2
x − 4= lim
x→2
√x − 2
(√
x − 2)(√
x + 2)
= limx→2
1√x + 2
=1
4
Example
Try limx→a
3√
x − 3√
a
x − a.
Finding limits by algebraic manipulations
Example
Find limx→4
√x − 2
x − 4.
SolutionWrite the denominator as x − 4 =
√x
2 − 4 = (√
x − 2)(√
x + 2).So
limx→2
√x − 2
x − 4= lim
x→2
√x − 2
(√
x − 2)(√
x + 2)
= limx→2
1√x + 2
=1
4
Example
Try limx→a
3√
x − 3√
a
x − a.
Finding limits by algebraic manipulations
Example
Find limx→4
√x − 2
x − 4.
SolutionWrite the denominator as x − 4 =
√x
2 − 4 = (√
x − 2)(√
x + 2).So
limx→2
√x − 2
x − 4= lim
x→2
√x − 2
(√
x − 2)(√
x + 2)
= limx→2
1√x + 2
=1
4
Example
Try limx→a
3√
x − 3√
a
x − a.
Two More Important Limit Theorems
TheoremIf 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 possiblyat a), and
limx→a
f (x) = limx→a
h(x) = L,
thenlimx→a
g(x) = L.
Math 1a - October 01, 2007.GWBMonday, Oct 1, 2007
Page12of13
We can use the Squeeze Theorem to make complicated limitssimple.
Example
Show that limx→0
x2 sin
(1
x
)= 0.
SolutionWe have for all x,
−x2 ≤ x2 sin
(1
x
)≤ x2
The left and right sides go to zero as x → 0.
We can use the Squeeze Theorem to make complicated limitssimple.
Example
Show that limx→0
x2 sin
(1
x
)= 0.
SolutionWe have for all x,
−x2 ≤ x2 sin
(1
x
)≤ x2
The left and right sides go to zero as x → 0.
Math 1a - October 01, 2007.GWBMonday, Oct 1, 2007
Page13of13
We can use the Squeeze Theorem to make complicated limitssimple.
Example
Show that limx→0
x2 sin
(1
x
)= 0.
SolutionWe have for all x,
−x2 ≤ x2 sin
(1
x
)≤ x2
The left and right sides go to zero as x → 0.
