limit and continuity student

7/30/2019 limit and continuity student

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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.

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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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limit and continuity student

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