d section 2.2 limits 2 lectures - abdulla eid...d to nd the limit lim x!a f (x), we have 1...
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Dr. AbdullaEid
Section 2.2Limits
2 Lectures
Dr. Abdulla Eid
Department of Mathematics
MATHS 101: Calculus I
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Definition of a limit
Example 1
How does the function
f (x) =x2 − 1
x − 1
behave at x = 1?
Solution:
f (1) =12 − 1
1− 1=
0
0undefined!
Hence, we cannot substitute directly with x = 1, so instead we checkvalues that are very much close to x = 1 and we check the correspondingvalues of f (x).
x 0.9 0.99 0.9999 1 1.00001 1.001 1.01
f (x) -
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Continue...
So we have seen that as x approaches 1, f (x) approaches 2, we write
limx→1
f (x) = 2
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Exercise 2
Using the method of the previous example (the table) find the followinglimits:
1 limx→5 x .
2 limx→a x .
3 limx→a 7.
4 limx→a k .
5 limx→0 f (x), where
f (x) =
{0, x < 0
1, x ≥ 0
Question: In order to find the limit of a function, do we need to do thetable method every time?
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Dr. AbdullaEidTo find the limit limx→a f (x), we have
1 Substitute directly by x = a in f (x). If you get a real number, thenthat is the limit.
2 If you get undefined values such as 00 , we use algebraic method to
clear any problem.
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Dr. AbdullaEid
Zero denominator of a rational functionA - Eliminating zero denominator by canceling common factor in thenumerator and denominator (00 form).
Example 3
Find
limx→5
x2 − 25
x − 5
Solution: Direct substitution gives
52 − 25
5− 5=
0
0undefined!
So we factor both the denominator and numerator to cancel the commonzero.
limx→5
x2 − 5
x − 5= lim
x→5
(x − 5)(x + 5)
(x − 5)
= limx→5
(x + 5)
= 5 + 5 = 10Dr. Abdulla Eid (University of Bahrain) Limits 6 / 26
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Exercise 4
Find
limx→1
x2 − 2x + 1
x − 1
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Example 5
Find
limx→3
x2 − 3x
x2 − 9
Solution: Direct substitution gives
02 − 3(0)
02 − 9=
0
0undefined!
So we factor both the denominator and numerator to cancel the commonzero.
limx→3
x2 − 3x
x2 − 9= lim
x→3
x(x − 3)
(x − 3)(x + 3)
= limx→3
x
x + 3
=3
6=
1
2
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Exercise 6
Find
limy→0
5y3 + 8y2
3y4 − 16y2
Solution: Direct substitution gives
5(0)3 + 8(0)2
3(0)4 − 16(0)2=
0
0undefined!
So we factor both the denominator and numerator to cancel the commonzero.
limy→0
5y3 + 8y2
3y4 − 16y2= lim
y→0
y2(5y + 8)
y2(3y2 − 16)
= limy→0
5y + 8
3y2 − 16
=5(0) + 8
3(0)2 − 16=
8
−16=−1
2
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Exercise 7
Find
limx→1
x2 − 3x + 4
x2 − 6x + 7
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Example 8
Find
limx→−1
2x2 + 3x + 1
x2 − 2x − 3
Solution: Direct substitution gives
2(0)2 + 3(0) + 1
(0)2 − 2(0)− 3=
0
0undefined!
So we factor both the denominator and numerator to cancel the commonzero.
limx→−1
2x2 + 3x + 1
x2 − 2x − 3= lim
x→−1
2(x + 12 )(x + 1)
(x − 3)(x + 1)
= limx→−1
2(x + 12 )
x − 3
=2(− 1
2 )
−4=
1
4
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Exercise 9
Find
limx→2
x3 − 8
3x2 − x − 10
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Example 10
Find
limx→1
4x5 − 4
5x2 − 5
Solution: Direct substitution gives
4(0)5 − 4
5(0)2 − 5=
0
0undefined!
So we factor both the denominator and numerator to cancel the commonzero.
limx→1
4x5 − 4
5x2 − 5= lim
x→1
4(x5 − 1)
5(x2 − 1)
= limx→1
4(x − 1)(x4 + x3 + x2 + x + 1)
5(x − 1)(x + 1)
= limx→1
4(x4 + x3 + x2 + x + 1)
5(x + 1)
=20
10= 2
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Example 11
Find
limx→1
√2
3
Solution: Direct substitution gives
√2
3
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Exercise 12
Find
limx→2
x4 − 16
x3 − 8
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Conjugate and multiply by 1
B - Eliminating zero denominator by multiplying with the conjugate.
Example 13
Find
limx→9
√x − 3
x − 9
Solution: Direct substitution gives
√9− 3
9− 9=
0
0undefined!
Since there is a square root and we cannot factor anything, we multiplyboth the numerator and denominator with the conjugate of the numerator(the one that contains the square root).
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Continue...
limx→9
√x − 3
x − 9= lim
x→9
(√x − 3)
(x − 9)
(√x+3)
(√x+3)
= limx→9
x − 9
(x − 9)(√x + 3)
= limx→9
1
(√x + 3)
=1
6
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Exercise 14
Find
limx→9
√x − 5− 2
x − 9
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Conjugate and multiply by 1
Example 15
Find
limx→7
√x + 2− 3
x2 − 49
Solution: Direct substitution gives
√9− 3
49− 49=
0
0undefined!
Since there is a square root and we cannot factor anything, we multiplyboth the numerator and denominator with the conjugate of the numerator(the one that contains the square root).
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Continue...
limx→7
√x + 2− 3
x2 − 49= lim
x→7
(√x + 2− 3)
(x2 − 49)
(√x + 2+3)
(√x + 2+3)
= limx→7
x + 2− 9
(x − 7)(x + 7)(√x + 2 + 3)
= limx→7
(x − 7)
(x − 7)(x + 7)(√x + 2 + 3)
= limx→7
1
(x + 7)(√x + 2 + 3)
=1
84
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Exercise 16
Find
limx→4
x − 4√x − 2
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Properties of Limits
Let limx→a f (x) = L and limx→a g(x) = K .
1 limx→a c= c .
2 limx→a xn = an.
3 limx→a [f (x) + g(x)] = limx→a f (x) + limx→a g(x) =L+K .
4 limx→a [f (x) · g(x)] = limx→a f (x) · limx→a g(x) =L ·K .
5 limx→a [cf (x)] = c limx→a f (x) =cL.
6 limx→a
[f (x)g (x)
]= limx→a f (x)
limx→a g (x)= L
K if K 6= 0.
7 limx→an√
f (x) = n√
limx→a f (x) = n√L. If n is even, then L must be
non–negative.
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Summary: These properties are telling us we can substitute directly withthe value of a if there no problem.
Exercise 17
Find
limx→−3
(x + 3
x2 − 9
)102
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The Sandwich Theorem (Squeeze theorem)
Theorem 18
Suppose g(x) ≤ f (x) ≤ h(x), i.e., f is squeezed between g and h.Suppose also that
limx→a
g(x) = limx→a
h(x) = L
Then we must havelimx→a
f (x) = L
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Example 19
Find limx→0 u(x) if 1− x2
4 ≤ u(x) ≤ 1 + x2
2 .
Solution:
1− x2
4≤ u(x) ≤ 1 +
x2
2
limx→0
1− x2
4≤ lim
x→0u(x) ≤ lim
x→01 +
x2
2
1 ≤ limx→0
u(x) ≤ 1
limx→0
u(x) = 1
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Exercise 20
Find limx→0 f (x) if√
5− 2x2 ≤ f (x) ≤√
5− x2.
Solution: √5− 2x2 ≤ f (x) ≤
√5− x2
limx→0
√5− 2x2 ≤ lim
x→0f (x) ≤ lim
x→0
√5− x2
√5 ≤ lim
x→0f (x) ≤
√5
limx→0
f (x) =√
5
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