Post on 04-Apr-2018
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Limit and Continuity
In this first lecture, we study the functions, that is the way in which one quantityy depends upon some other quantity x, according to y = f(x). In particular, we shallstudy those functions for which y varies continuously with x, in the sense that a small
change in x results in a small change in y. First, let us consider some examples:
Example A
x
y
f(x) = x2
Example B
x
y
f(x) =
x2 , x = 01 , x = 0
Example C
x
y
f(x) =
1 , x 01 , x < 0
The function in Example A should certainly be considered to be continuous. Thereare no breaks or jumps in its graph, and it is clear that a small change in x producessmall change in x2. In Example B, at x = 0, y = 1, a slightly change in x away from 0will result in a sudden jump in the value of f(x) to a value near 0. The graph is brokenat x = 0, so that Example B is not a continuous function. Example C illustrates afunction which is not continuous because of the jump at the origin.
In this lecture we study the behavior of f(x) when x approaches a value, say, a.
approach does not mean equal Two ways to approach a given value:
(i) approach from right, (ii) approach from left.
10 2
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From right 1.5 1.1 1.01 1.001From left 0.5 0.9 0.99 0.999
The arrow on the left points to the right in the above diagram indicates that x ap-proaches the value 1 from the left. Likewise, the arrow on the right points to the left
indicates that x approaches the value 1 from the right.
Example 1
Consider the function f(x) = 2x + 1, what happen to f(x) when x get closer and closerto 3?
x f(x)
2.5 62.9 6.08
2.99 6.892.999 6.998
Table 1: f(x) approaches 7 when x get closer to 3 from left.x f(x)
3.1 7.23.01 7.02
3.001 7.0023.0001 7.0002
Table 2: f(x) approaches 7 when x get closer to 3 from right.
The table in above shows that the function f(x) approaches 7 when x get closer andcloser to 3 from left and right.
Definitions:
Left-Hand Limit (LHL)
If f(x) gets closer to L as x approaches a from the left, we say that L is the left-handlimit of f(x) at a and we write
limxa
f(x) = L or limxa,x
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Iff(x) gets closer to L as x approaches a from the right, we say that L is the right-handlimit of f(x) at a and we write
limxa+
f(x) = L or limxa,x>a
f(x) = L
The symbol x
a+ indicates that we consider only values ofx that are greater than a.
Example 2
Considerg(x) = x2
What happen to g(x) when x gets closer and closer to 3?
32.9
x g(x) = x2
2.9 8.412.99 8.9401
2.9999 8.99942.9999999 8.999999
limx3
g(x) = 9
3 3.1
x g(x) = x2
3.1 9.613.01 9.0601
3.0001 9.00060013.0000001 9.0000006
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limx3+
g(x) = 9
As x gets closer to 3 separately from left and right, the function g(x) will separatelyapproaches 9. Therefore, the limit ofg(x) = x2 as x approaches 3 from either sides is 9.This is illustrated in the diagram below:
x
g(x)
3
9
Example 3
Consider
f(x) =
1 , x 01 , x > 0
What happen to f(x) when x gets closer and closer to 0?
x f(x)-0.5 -1-0.1 -1
-0.01 -1-0.001 -1
As x approaching 0 from left we see that f(x) approaches -1, i.e.
limx0
f(x) = 1
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x f(x)0.5 10.1 1
0.01 10.001 1
As x approaching 0 from right we see that f(x) approaches 1, i.e.
limx0+
f(x) = 1
0
1
1
x
f(x)
The limits of f(x) from the left and the right are not equal, thus we may concludethat LIMIT DOES NOT EXIST as x 0.
Question: When does the limit exist?
The limit of a function exists when
limxa
f(x) = L limxa+
f(x) = L
equal
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that is to say that the left-hand limit of f(x) as x approaches a from the left is equal tothe right-hand limit of f(x) as x approaches a from the right. This can be written as
limxa
f(x) = L
This means that we make the values of f(x) arbitrarily close to L by taking x to besufficiently close to a (on either side ofa) but x = a. In other word, in finding the limitof f(x) as x gets closer and closer to a, we never consider x = a. In most of the cases,f(x) need not even be defined when x = a. The thing that we only concern is how f(x)is defined near a regardless of what happens at a. Conversely, if
limxa+
f(x) = R and limxa
f(x) = L
where R = L, then the limxa f(x) does not exist.
Example 4
Evaluate the right-hand and left-hand limits of f(x) = x2 + 2x 1 at x = 1.
x f(x) = x2 + 2x 11.1 2.41
1.01 2.04011.0001 2.000400001
1.000001 2.000004000001
Thus, we have
limx1+ f(x) = 2
x f(x) = x2 + 2x 10.9 1.61
0.99 1.96010.999 1.996001
0.9999 1.99960001
Thus, we havelimx1
f(x) = 2
Since LHL and RHL are equal
limx1
f(x) = 2 and limx1+
f(x) = 2
Therefore, the limit of f(x) does exist and can be written as
limx1
f(x) = 2
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Example 5
f(x) =
2x 1 , x < 21
x, x 2
What happen when x gets closer and closer to 2?From left:
x f(x)1.9 2.8
1.99 2.981.999 2.998
limx2
f(x) = 3
From right:
x f(x)2.01 0.49752.001 0.4996
2.0001 0.49998
limx2+
f(x) = 0.5
Thus,limx2
f(x) = 3 and limx2+
f(x) = 0.5
Sincelimx2
f(x) = limx2+
f(x)
Therefore, the limit off(x) does not exist. The diagram of function f(x) is shown below:
f(x)
3
0.5
1x2
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Continuity
Using function notation we represent the value of the function f(x) at x = a asf(a). Function notation gives us a nice compact way of representing function values.To evaluate the function, everywhere we see an x on the right side we will substitute
whatever is in the parenthesis on the left side. For example, to evaluate the functionf(x) = 2x2 5x + 4 at x = 1, we get
f(1) = 2(1)2 5(1) + 4 = 1
This is represented by the diagram below:
0
1
2
0 1 2x
y
(1, f(1))
Let us consider the following function:
f(x) =
x2 + x + 4 , x < 2 ,1
2x + 1 , x > 2 ,
5 , x = 2 .
It can be shown that (by evaluating the LHL and RHL) the limit of f(x) exists when xapproaches 2,
limx2
f(x) = 2
However, f(2) = 5. In this case,
limx2
f(x) = f(2) .
This function is represented by the graph below:
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0
1
2
3
4
5
0 1 2 3 4 5x
y
This curve is discontinuous because there is a sudden jump/broken when x = 2.
Definition 1:
A function f is continuous at a number a if
limxa
f(x) = f(a)
This requires three conditions if f is continuous at a:
1. f(a) is defined (i.e., a is in the domain of f)
2. limxa
f(x) exists
3. limxa
f(x) = f(a)
These statements say that f(x) is continuous at a iff(x) approaches f(a) as x approachesa. Likewise, if f(x) is defined on an open interval containing a (except at a), we saythat f(x) is discontinuous at a if f(x) is not continuous at a.
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Back to the example above, if f(2) = 2, the curve becomes
0
1
2
3
4
5
0 1 2 3 4 5
x
y
The curve is now continuous as the broken point on the curve has been filled by thesolid dot.
Definition 2:
A function f is continuous from the right at a number a if
limxa+ f(x) = f(a)
(See Example 5 of the previous section.)
A function f is continuous from the left at a number a iflimxa
f(x) = f(a)
(See Example 3 of the previous section.)
Definition 3:
A function f is continuous on an interval if it is continuous at every number in theinterval.
Example 1
Explain why each of the following functions are discontinuous?
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1. f(x) =x2 x 2
x 2
2. f(x) =
1
x2, x = 0
1 , x = 0
3. f(x) =
x2 x 2x 2 , x = 2
1 , x = 2
4. f(x) = [[x]]
Solution:
1. Condition 1 fails because f(2) is not defined.
2. Condition 2 fails because limx0 f(x) does not exist (infinity).
3. Condition 3 fails because limx2
f(x) = f(2).
4. Condition 2 fails because limxn+
f(x) = n but limxn
f(x) = n1, thus limxn
f(x) does
not exist.
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Evaluate limit of functions
There are three ways to evaluate limits
1. Substitution
2. Factorization
3. Conjugate
Substitution method
Example 1
Given f(t) =sin2t
t, using substitution method evaluate lim
tf(t).
Solution:
limt
f(t) = limt
sin2t
t
= limt
sin2()
=0
= 0
Theorem 1:
A polynomialP(x) = c0 + c1x + + cn1xn1 + cnxn
is continuous everywhere, i.e., it is continuous on R = (= , ).Remark: For all continuous functions, we can evaluate limit using Substitution
method.
Definitions:
For any polynomial functions
P(x) = c0 + c1x + c2x2 + + cnxnwhere c0, c1, c2, are constant. For any real number of a,
limxa
P(x) = limxa
(c0 + c1x + c2x2 + + cnxn)
= c0 + c1a + c2a2 + + cnan
= P(a)
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Example 2
Given f(x) = x2 x 2, using substitution method evaluate limx2
f(x).
Solution:
limx2
f(x) = limx2
(x2 x 2)= (2)2 (2) 2= 0
Definitions:
A rational function f(x) is a ratio of two polynomials
f(x) = P(x)Q(x)
, Q(x) = 0
where P(x) and Q(x) are polynomials.
Theorem 2:
Any rational function
f(x) =P(x)
Q(x), Q(x) = 0
is continuous wherever it is defined (in its domain).
Example 3
Evaluate
limx2
2x3 3x2 + 25x 3
Solution:
This function is rational, by Theorem 2 that it is continuous on its domain exceptat x = 3
5. Therefore
limx2
2x3 3x2 + 25x 3 =
2(2)3 3(2)2 + 25(2) 3 =
6
7.
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Example 4
Given f(x) =x2 x 2
x 2 , evaluate limx2 f(x).Solution:
What happen when we use the Substitution method? If we substitute x = 2 intof(x) we obtain a value which is equal to zero divided by zero.
Important Remarks:
The substitution method cannot be used if the denominator of a rational functiongives 0.
Factorization Method
If a rational function is formed by some polynomial functions and if it is factorizable,
we can use the factorization method to solve the problem (Example 4 continued):
limx2
x2 x 2x 2 = limx2
(x 2)(x + 1)x 2
= limx2
(x + 1) = 3
Conjugate Method
Quick Review:
Recall what is conjugate, for examplex 3 conjugate x + 3
x 1 + 2 conjugate
x 1 2
Example 5
Evaluate
limx4
x 4x 2
Solution:
As the function consists of square root, we can use the conjugate method to solvethis problem as follows:
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limx4
x 4x 2 = limx4
x 4x 2
x + 2x + 2
= limx4
(x 4)(x + 2)x
4
= limx4
(x + 2)=
4 + 2 = 4
Example 6
Evaluate
limx3
3 xx + 1 2
Solution:
limx3
3 xx + 1 2 = limx3
3 xx + 1 2
x + 1 + 2x + 1 + 2
= limx3
(3 x)(x + 1 + 2)x + 1 4
= limx3
(3 x)(x + 1 + 2)x 3
= limx3
(x 3)(x + 1 + 2)x 3
= limx3
(x + 1 2) = 4
EXERCISES 1:
Evaluate each limit using the suitable method.
1. limx2
x 1x2 x 1
2. limx1
x 1x2 1
3. limx2
x
1
1
x 2
4. limx1/2
2x + 1
2x2 x 1
5. limx2
x2 x 2x2 3x + 2
6. limx1
x2
1
2 x x7. Given
f(x) =
x + 2 , x < 1
x2 + 4 , x 1Sketch the graph for the function f(x), and find lim
x1f(x).
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Infinite Limit and Limit at Infinity
Definitions:
Let f be any function defined on both sides of a, except possibly at a itself. Iflimxa f(x) = , this means that the value of f increases (positive) or decreases(negative) without bound as x gets closer and closer to a, but not equal to a.
It is important to note that the is not regarded as a number. It just to express thevalues of f(x) do not approach a number, i.e., the values f(x) become larger and larger(positive or negative) as x becomes closer and closer to a, and so limxa f(x) does notexist.
Example 1
Consider h(t) = 2t 1. What happen to h(t) as t gets closer and closer to 1?
limt1
h(t) = limt1
2
t 1Does any methods learn earlier work in this case? How to evaluate the limit of functionh(t) as t gets closer and closer to 1? In fact, it appears from the table of the function h(t)as shown below that the denominator t 1 is a small negative number as t approaches 1from the left, and h(t) is numerically large negative. Likewise, the denominator t 1 isa small positive number as t approaches 1 from the right, and h(t) is numerically large
positive.
t h(t) t h(t)0.9 -20 1.1 20
0.999 -2000 1.001 20000.99999 -200000 1.00001 200000
limt1
2
t 1 = limt1+2
t 1 = +
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Lets sketch the graph to see how this function behaves.
t
h(t)
0 1
2
t 1
We see above that LHL and RHL at t 1 are not equal and thus the limit of h(t)does not exist.
Example 2
Evaluate
limx1
2
(x 1)2
Solution:
x2
(x
1)2
x2
(x
1)2
0.8 50 1.1 500.9 200 1.01 200
0.99 20000 1.001 200000.999 2000000 1.0001 2000000
limx1
2
(x 1)2 = + limx1+2
(x 1)2 = +
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As x gets closer to 1, both the LHL and RHL are approaching positive infinity. Thus,the values of the function do not approach a number, so the limit does not exist.
Definitions:
Vertical Asymptote
The line x = a is called a vertical asymptote of the curve y = f(x) if at least one ofthe following statements is true:
1. limxa
f(x) = 2. lim
xaf(x) =
3. limxa+
f(x) = 4. lim
xa
f(x) =
5. limxa
f(x) = 6. lim
xa+f(x) =
For instance, it can be seen in the graph in page 17 that the line t = 1 is a verticalasymptote of the curve h(t) = 2/(t 1).
Horizontal Asymptote
The line y = L is called the horizontal asymptote of the curve y = f(x) if either
limx
f(x) = L or limx
f(x) = L
is true.
Example 3
x 1/x0.1 10
0.01 1000.001 1000
0.0001 10000
limx0, x>0
1x
=
The y-axis is a vertical asymptote of the curve 1/x since it satisfies condition 3.
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Example 4
limx1+
x + 1x2 1 = limx1+
(x 1)(x 1)(x + 1)
= limx1+
1x + 1 =
The line x = 1 is a vertical asymptote.
Example 5
Considerf(x) =
x
x 1What happen to f(x) when x gets bigger and bigger and bigger.. towards INFINITY?
xx
x 1101 1.01
1001 1.00110001 1.0001
100001 1.00001
Thus,
limx
x
x
1
= 1
In this case, the line y = 1 is the horizontal asymptote.
Example 6
Consider
limx
2
x + 1
What happen to f(x) when x is large?
x2
x + 199 0.02
999 0.0029999 0.0002
99999 0.00002
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Thus,
limx
2
x + 1= 0
The curve 2/(x + 1) approaches 0 as x getting larger and larger. The illustration of thiscase is given in the diagram below:
x01
Remark: At infinity the limit of a polynomial is given by the limit of the monomialwith the highest degree.
Example 7
limx
(2x2 x 1) = limx
2x2
1 x2x2
12x2
= limx
2x2
1 12x
12x2
= lim
x2x2 =
Example 8
Evaluate
limx
2x3 5x2 3x3 1
What is the horizontal asymptote?
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Solution:
limx
2x3 5x2 3x3 1 = limx
x3
2 5x2
3x3
x3
1 1x3
= lim
x2
5
x2 3
x31 1x3 = 2
Now, we consider the highest degree terms.
limx
2x3 5x2 3x3 1 = limx
2x3
x3= 2
Thus, y = 2 is the horizontal asymptote.
Example 9
Evaluate
limx
9x4 5x2 3
3x3 1What is the horizontal asymptote?
Solution:
limx
9x4 5x2 3
3x3 1 = limx
9x4
3x3
= limx
3x2
3x3
= limx
1
x = 0
Thus, y = 0 is the horizontal asymptote.
EXERCISE 2:
Evaluate the limit and find the vertical/horizontal asymptote.
1. limx2+
x
x2 4
2. limx
xx2 4
3. limx
2x2 + 3x 13x2 2x + 4
4. Find limxa
C, where a and C are constants.
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Rules and Theorems
Rules:
(I) Finite case
Let a, b and c be any arbitrary real numbers, and suppose that
limxa
f(x) = L1 and limxa
g(x) = L2
Then
1. limxa
[f(x) + g(x)] = L1 + L2
2. limxa
[f(x) g(x)] = L1 L2
3. limxa[cf(x)] = cL1 , for any c R
4. limxa
[f(x)g(x)] = L1L2
5. limxa
f(x)
g(x)
=
L1L2
, if L2 = 0
6. limxa
[f(x)]n = L1n
For those who are interested to verify the rules in above, please refer to the appendix.
Example 1
Givenlimt2
F(t) = 8 , limt2
G(t) = 2 , limt2
H(t) = 0.1
Compute
(i) limt2
[2F(t)G(t)H(t)]
(ii) limt2
3
F(t)
(iii) lim
t2
F(t)G(t)
+ lim
t2[H(t)]2
Solution:
(i) limt2
[2F(t)G(t)H(t)] = 2(8)(2)(0.1) = 3.2
(ii) limt2
3
F(t)
=3
8 = 2
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(iii) limt2
F(t)
G(t)
+ lim
t2[H(t)]2 =
8
2 + (0.1)2 = 3.99
(II) Infinite case
When we evaluate the limit of a function, we have to find an appropriate method if the
following indeterminate forms are encountered:
1. infinity minus infinity, 2. infinity times zero, 03. infinity divided by infinity,
4. zero divided by zero, 00
Example 2
Givenlimt2
F(t) = , limt2
G(t) = 2 , limt2
H(t) = 0 .
Compute
(i) limt2
[2F(t)G(t)]
(ii) limt2
3
F(t)
(iii) limt2
F(t)G(t)
+ limt2
G(t)F(t)
2
Solution:
(i) limt2
[2F(t)G(t)] =
(ii) limt2
3
F(t)
=
(iii) limt2
F(t)
G(t)
+ lim
t2
G(t)
F(t)
2=
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Example 3
Evaluatelimt
(t2 t) =?
Solution:Since
limt
t2 = and
limt
t = .So that
limt
(t2 t) = is an indeterminate form. But with different approach we can
limt
(t2 t) = limt
t(t 1) = limt
t(t) = limt
t2 =
Theorem 3:
Iff(x) g(x) when x is close to a (except possibly at a) and the limits off and g bothexist as x approaches a, then
limxa
f(x) limxa
g(x)
Theorem 4: The Squeeze Theorem
If f(x) g(x) h(x) when x is close to a (except possibly at a) andlimxa
f(x) = limxa
h(x) = L
thenlimxa
g(x) = L
The Squeeze Theorem says that if f(x) and h(x) have the same limit L at a, then g(x)is forced to have the limit L at a.
Example 4
Evaluate
limx0
x2 sin1
x
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Solution:
From our trigonometric knowledge
1 sin
1
x 1
x2 x2 sin1x x2Since
limx0
(x2) = limx0
x2 = 0
By using Theorem 3,
x2 x2 sin
1
x
x2
limx0
(x2) limx0
x2 sin
1
x lim
x0x2
0 limx0
x2 sin
1x 0
From Theorem 4 we conclude that
limx0
x2 sin1
x
= 0
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Appendix
Note: This section is for reference only.
Proof of rule 1 (Sum Rule):
Let > 0 and consider |[f(x) + g(x)] (L1 + L2)|. By the triangle inequality that
|[f(x) + g(x)] (L1 + L2)| |f(x) L1| + |g(x) L2| .
Now choose 1 > 0 and 2 > 0 so that
0 < |x a| < 1 |f(x) L1| < /2 ,and 0 < |x a| < 2 |g(x) L2| < /2 .
Next put = min(1, 2). If 0 < |x a| < , both conditions above come into operation and so
|[f(x) + g(x)] (L1 + L2)| |f(x) L1| + |g(x) L2| < /2 + /2 = ,
i.e., f(x) + g(x) L1 + L2, as required.
Proof of Rule 3 (Coefficient Rule):
If c = 0, the function cf(x) is just the constant zero function and the result is clear. So assumec = 0. Let > 0. Choose > 0 such that
0 < |x a| < |f(x) L1| < |c| .
It follows that
|c f(x) cL1| = |c| |f(x) L1| < |c|
|c| = ,i.e., 0 < |x a| < |(cf)(x) cL1| < , as required.
Proof of Rule 2 (Difference Rule):
By using Rule 3,g(x) = (1) g(x) 1 L2 = L2 .
Then, by using Rule 1,
f(x) g(x) = f(x) + g(x) L1 + (L2) = L1 L2 .
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Proof of Rule 4 (Product Rule):
Writef(x)g(x) L1L2 = [f(x) L1]L2 + f(x) [g(x) L2] .
Then
|f(x)g(x)
L1L2
| |f(x)
L1
||L2
|+
|f(x)
| |g(x)
L2
|.
Since limxa f(x) = L1, there exists > 0 such that 0 < |x a| < |f(x) L1| < 1, which in turnimplies that |f(x)| |f(x)L1|+ |L1| < 1 + |L1|. So 0 < |xa| < |f(x)| < K, where K = 1 + |L1|.Combining with the earlier inequality shows that if 0 < |x a| < , then
0 |f(x)g(x) L1L2| |f(x) L1||L2| + K |g(x) L2| .
As x a, |f(x) L1| 0 and so |f(x) L1||L2| 0, by Rule 3. Similarly, |L1| < 1 + |L1| 0 asx a. Therefore, by Rule 1, the right-hand side of inequality above tends to 0, as x a. It followsby the squeeze theorem that |f(x)g(x) L1L2| 0 as x a, i.e., f(x)g(x) L1L2.
Proof of Rule 5 (Quotient Rule):
We write 1L2 1g(x)
= |g(x) L2||L2| |g(x)| .There exist > 0 such that 0 < |x a| < |g(x)| > |L2|/2. For such values of x then
0 1L2 1g(x)
2|L2|2 |g(x) L2| .As x a the right-hand side of this inequality tends to 0 (by using Rule 3). So by the squeeze theorem
1
L2 1
g(x)
0 , i.e., 1
g(x) 1
L2.
From this result and Rule 4,f(x)
g(x)= f(x) 1
g(x) L1 1
L2=
L1L2
,
as required.
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