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# Also known as Hysterical Calculus

## 1.

Calculus++ Light## 2.

Question 1. A sequence an, n = 1,2,3,…,satisfies lim (2n 1)an 16 .

n

a) Use the definition of limit to obtain a

sandwich inequality for an.

Solution: Since the limit of (2n – 1) an is 16 we

have: 0, N , n N : | (2n 1)an 16 | .

Set 1, then | (2n 1)an 16 | 1, n N .

That is, 1 (2n 1)an 16 1

15 (2n 1)an 17

15

17

an

lim an 0.

n

2n 1

2n 1

## 3.

b) Conclude that lim an 0,n

and find lim nan .

n

We have nan 12 (2n 1)an 12 an

Therefore

lim nan lim 12 (2n 1)an lim 12 an

n

n

n

12 lim (2n 1)an 12 lim an

n

12 16 12 0 8.

n

## 4.

Calculus++Also known as

Hysterical Calculus

## 5.

Question 2. A sequence xn, n = 1,2,3,… isxn 1 xn 2

defined by the relationship xn

2

and the initial conditions x1 = a, x2 = b.

Find lim xn .

n

Solution. We begin with finding an explicit

expression for the general term of the

sequence xn.

n

Let us try the following formula: xn c .

xn 1 xn 2

n

n 1

n 2

1

1

xn

c 2 c 2 c .

2

2

n 2

Divide both sides by c to obtain 12 12 ,

2

or 12 12 0 1 1, 2 12 .

## 6.

Thus, we found two sequences that satisfy thexn 1 xn 2

defining relationship xn

:

2

( 2)

(1)

1 n

xn c1 and xn c2 ( 2 ) .

Do any of these sequences satisfy the initial

conditions x1 = a, x2 = b?

Well, if a = b, then the first sequence with c1 = a,

(1)

xn a, satisfies the initial conditions.

If b = – ½ a, then the second sequence with

c2 = –2 a, xn( 2 ) a ( 12 ) n 1 , satisfies the initial

conditions.

But what should we do if a and b are arbitrary?

## 7.

Well, we can consider linear combination of1 n

the two obtained sequences xn c1 c2 ( 2 ) .

Let us check that this linear combination

xn 1 xn 2

.

indeed satisfies the equation xn

2

We have

1 n 1

2

xn 1 xn 2 c1 c2 ( )

2

1 n 2

2

c1 c2 ( )

2

1 1

1 2

( 2 ) ( 2 )

1 n

c1 c2 ( 2 )

2

1 n

c1 c2 ( 2 ) xn .

## 8.

Now all we have to do is to find the values ofc1 and c2 such that our sequence also

satisfies the initial conditions:

x1 c1 c2 ( 12 ) a,

1

x2 c1 4 c2 b.

For the values of arbitrary constants c1 and c2

we obtain 3c1 a 2 b , 43 c2 b a.

a 2b 4

1 n

Thus xn

( b a )( 2 ) .

3

3

Now the limit n is not difficult to find:

1

2

lim xn 3 a 3 b.

n

## 9. The method of the week

To find the sequence that satisfies the definingrelationship xn xn 1 xn 2 ,

and the initial conditions x1 = a, x2 = b we

have to:

1. Write down the characteristic equation

0

and obtain its roots 1 , 2 .

2

2. Write down the general formula for xn:

xn c c ,

n

1 1

n

2 2

and find the values of constants c1 and c2,

such that x1 = a, x2 = b.

## 10.

Question 3 a). Find the following limitSolution: We have

sin n 1 sin n 1 n n

sin n 1 n cos( n )

cos n 1 n sin( n )

0

sin n 1 n ( 1) .

lim sin n 1 .

2

n

2

2

2

2

2

n

## 11.

n 1 n2

2

2

n 1 n

n 1 n

2

n 1 n

2

2

n 1 n

n 1 n

1

.

2

n 1 n

2

We have

The obtained identity yields

sin n 1 sin n 1 n ( 1)

2

2

n

sin

(

1

)

.

2

n 1 n

n

## 12.

Therefore we can use the following sandwichinequality

2

sin

sin

n

1

2

n 1 n

.

sin

2

n 1 n

Since sin x is a continuous function we obtain

sin lim

lim sin

n 2

2

n

n 1 n

n 1 n

sin 0 0 .

Hence, the sandwich theorem tells us that

lim sin n 1 0 .

n

2

## 13.

Question 4. State a (positive) definition of adivergent sequence {xn}.

Solution: We begin with the definition of a

convergent sequence.

A sequence {xn} converges to a number L, if

0, N , n N : | x n L | .

A sequence {xn} does not converges to a

number L, if

0, N , n N : | x n L | .

A sequence {xn} is divergent, if it does not

converges to any number L.

L, 0, N , n N : | xn L | .

## 14.

Question 5. Draw the curve defined by then

n

n

equation lim | x | | y | 1

n

in the x y – plane.

Solution. To begin with, we calculate the limit

lim | x | | y | ,

n

n

n

n

in the particular case x = – 7, y = 5.

We have | 7 |n | 7 |n | 5 |n 2 | 7 |n .

| 7 | | 7 | | 5 | 2 | 7 | .

Since lim n 2 1, the sandwich theorem

n

n

n

n

tells us that lim | 7 | | 5 | | 7 | 7 .

Hence,

n

n

n

n

n

## 15.

Now we can find the limit limn

n

|x| |y| .

n

n

Note the following double inequality

max(| x |, | y |)

Hence

n

|x| |y|

n

2 max(| x |, | y |) .

n

n

max(| x |, | y |) | x | | y |

n

n

n

2 max(| x |, | y |).

n

## 16.

Since lim 2 1, the sandwich theoremn

n

tells us that

lim | x | | y | max(| x |, | y |).

n

n

n

n

Thus, we have to draw the curve defined by

the equation

max(| x |, | y |) 1 .

## 17.

Let us look at the xy – plane:y x

max(| x |, | y |) y

y

y 1

1

x 1

max(| x |, | y |) x

y x

x 1

max(| x |, | y |) x

x

1

1

y 1 1

max(| x |, | y |) y

## 18.

The graph of the curve limy x

n

y 1

y

1

x 1

n

| x | | y | 1.

y x

n

n

x 1

x

1

1

y 1

1

## 19.

Question 6. Use the definition of convergentsequence to obtain a sandwich inequality

for the sequence

n

1007 sin n

xn

, n 1,2,3,...,

n

1008

and find the limit of this sequence.

sin n

Solution: The sequence y n

n

converges to 0.

Therefore, according to the definition of the

limit, 0, N , n N : | y n 0 | .

1

and denote N1 – the

Choose

2016 corresponding value of N.

## 20.

The definition tells us that1

sin n

1

for all n N 1 .

2016

n

2016

1007

1

1007 sin n 1007

1

1008 2016 1008

n

1008 2016

2013 1007 sin n 2015

2016 1008

n

2016

Therefore we obtain the following sandwich

inequality for our sequence xn

n

n

n

0

2013 1007 sin n 2015

n 2016

2016 1008

0

for all n N 1 .

## 21.

Now the sandwich theorem tells us thatn

1007 sin n

lim

0.

n 1008

n