A Neutrosophic Binomial Factorial Theorem with their Refrains

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Neutrosophic Sets and Systems, Vol. 14, 2016

University of New Mexico

A Neutrosophic Binomial Factorial Theorem with their Refrains Huda E. Khalid1 1

Florentin Smarandache2

Ahmed K. Essa3

University of Telafer, Head of Math. Depart., College of Basic Education, Telafer, Mosul, Iraq. E-mail: hodaesmail@yahoo.com 2 University of New Mexico, 705 Gurley Ave., Gallup, NM 87301, USA. E-mail: smarand@unm.edu 3 University of Telafer, Math. Depart., College of Basic Education, Telafer, Mosul, Iraq. E-mail: ahmed.ahhu@gmail.com

Abstract. The Neutrosophic Precalculus and the Neutrosophic Calculus can be developed in many ways, depending on the types of indeterminacy one has and on the method used to deal with such indeterminacy. This article is innovative since the form of neutrosophic binomial factorial theorem was constructed in addition to its refrains.

Two other important theorems were proven with their corollaries, and numerical examples as well. As a conjecture, we use ten (indeterminate) forms in neutrosophic calculus taking an important role in limits. To serve article's aim, some important questions had been answered.

Keyword: Neutrosophic Calculus, Binomial Factorial Theorem, Neutrosophic Limits, Indeterminate forms in Neutrosophic Logic, Indeterminate forms in Classical Logic.

1 Introduction (Important questions)

2. Let √đ??ź = đ?‘Ľ + đ?‘Ś đ??ź 0 + đ??ź = đ?‘Ľ 3 + 3đ?‘Ľ 2 đ?‘Ś đ??ź + 3đ?‘Ľđ?‘Ś 2 đ??ź2 + đ?‘Ś 3 đ??ź3 0 + đ??ź = đ?‘Ľ 3 + (3đ?‘Ľ 2 đ?‘Ś + 3đ?‘Ľđ?‘Ś 2 + đ?‘Ś 3 )đ??ź 3

Q 1 What are the types of indeterminacy?

There exist two types of indeterminacy a. Literal indeterminacy (I).

As example: 2 + 3đ??ź

3

đ?‘Ľ = 0, đ?‘Ś = 1 → √đ??ź = đ??ź.

(1)

√đ??ź = đ??ź,

(2)

Q 2 What is the values of đ??ź to the rational power?

1. Let

(3)

In general, 2đ?‘˜

√đ??ź = Âąđ??ź

(6)

where đ?‘˜ ∈ đ?‘§ + = {1,2,3, ‌ }.

meaning that the indeterminacy membership = 0.3.

Other examples for the indeterminacy component can be seen in functions: đ?‘“(0) = 7 đ?‘œđ?‘&#x; 9 or đ?‘“(0 đ?‘œđ?‘&#x; 1) = 5 or đ?‘“(đ?‘Ľ) = [0.2, 0.3] đ?‘Ľ 2 ‌ etc.

√đ??ź = đ?‘Ľ + đ?‘Ś đ??ź 0 + đ??ź = đ?‘Ľ 2 + (2đ?‘Ľđ?‘Ś + đ?‘Ś 2 )đ??ź đ?‘Ľ = 0, đ?‘Ś = Âą1.

In general, 2đ?‘˜+1

b. Numerical indeterminacy.

As example: đ?‘Ľ(0.6,0.3,0.4) ∈ đ??´,

(5)

(4)

Basic Notes 1. A component I to the zero power is undefined value, (i.e. đ??ź0 is undefined), đ??ź since đ??ź0 = đ??ź1+(−1) = đ??ź1 ∗ đ??źâˆ’1 = which is đ??ź

2.

impossible case (avoid to divide by đ??ź).

The value of đ??ź to the negative power is undefined value (i.e. đ??ź −đ?‘› , đ?‘› > 0 is undefined).

Q 3 What are the indeterminacy forms in neutrosophic calculus?

In classical calculus, the indeterminate forms are [4]: 0 0

∞

, , 0 ∙ ∞ , ∞0 , 00 , 1∞ , ∞ − ∞. ∞

(7)

+

where đ?‘˜ ∈ đ?‘§ = {1,2,3, ‌ }.

Huda E. Khalid, Florentin Smarandache & Ahmed K. Essa, A Neutrosophic Binomial Factorial Theorem with their Refrains

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Neutrosophic Sets and Systems, Vol. 14, 2016

The form 0 to the power đ??ź (i.e. 0đ??ź ) is an indeterminate form in Neutrosophic calculus; it is tempting to argue that an indeterminate form of type 0đ??ź has zero value since "zero to any power is zero". However, this is fallacious, since 0đ??ź is not a power of number, but rather a statement about limits.

Q 4 What about the form 1đ??ź ?

The base "one" pushes the form 1đ??ź to one while the power đ??ź pushes the form 1đ??ź to I, so 1đ??ź is an indeterminate form in neutrosophic calculus.

Indeed, the form đ?‘Žđ??ź, đ?‘Ž ∈ đ?‘… is always an indeterminate form.

Q 5 What is the value of đ?‘Žđ??ź , đ?‘¤â„Žđ?‘’đ?‘&#x;đ?‘’ đ?‘Ž ∈ đ?‘…?

Let đ?‘Ś1 = 2đ?‘Ľ , đ?‘Ľ ∈ đ?‘… , đ?‘Ś2 = 2đ??ź ; it is obvious that lim 2đ?‘Ľ = ∞ , lim 2đ?‘Ľ = 0 , lim 2đ?‘Ľ = 1; while đ?‘Ľâ†’∞

đ?‘Ľâ†’−∞

đ?‘Ľâ†’0 đ??ź

we cannot determine if 2 → ∞ đ?‘œđ?‘&#x; 0 đ?‘œđ?‘&#x; 1, therefore we can say that đ?‘Ś2 = 2đ??ź indeterminate form in Neutrosophic calculus. The same for đ?‘Žđ??ź , where đ?‘Ž ∈ đ?‘… [2].

[2.1, 2.5] [2.1, 2.5] = 1 1 ln đ?‘Ľ ln 0 [2.1, 2.5] [2.1, 2.5] = = 1 −0 −∞ 2.1 2.5 = [ , ] = (−∞, −∞) −0 −0 = −∞ Hence đ?‘Ś = đ?‘’ −∞ = 0 OR it can be solved briefly by đ?‘Ś = đ?‘Ľ [2.1,2.5] = [02.1 , 02.5 ] = [0,0] = 0. ∴ lim ln đ?‘Ś = lim đ?‘Ľâ†’0

�→0

Example (3.2) lim [3.5,5.9]đ?‘Ľ [1,2] = [3.5,5.9] [9,11][1,2] =

�→[9,11]

[3.5,5.9] [91 , 112 ] = [(3.5)(9), (5.9)(121)] = [31.5,713.9].

Example (3.3) lim [3.5,5.9] đ?‘Ľ [1,2] = [3.5,5.9] ∞[1,2] đ?‘Ľâ†’∞

= [3.5,5.9] [∞1 , ∞2 ] = [3.5 ∙ (∞) ,5.9 ∙ (∞)] = (∞, ∞) = ∞.

2 Indeterminate forms in Neutrosophic Logic

Example (3.4) Find the following limit using more than one

It is obvious that there are seven types of indeterminate forms in classical calculus [4],

technique lim

0 ∞

, , 0. ∞, 00 , ∞0 , 1∞ , ∞ − ∞.

0 ∞

As a conjecture, we can say that there are ten forms of the indeterminate forms in Neutrosophic calculus đ??ź ∞ đ??ź0 , 0đ??ź , , đ??ź ∙ ∞, , ∞đ??ź , đ??źâˆž , đ??źđ??ź , 0 đ??ź đ?‘Žđ??ź (đ?‘Ž ∈ đ?‘…), ∞ Âą đ?‘Ž ∙ đ??ź .

�→0

đ??ź

0

3 Various Examples Numerical examples on neutrosophic limits would be necessary to demonstrate the aims of this work.

Example (3.1) [1], [3] The neutrosophic (numerical indeterminate) values can be seen in the following function: Find lim �(�), where �(�) = � [2.1,2.5] . �→0

.

Using L'HĂ´pital's rule 1 −1 lim ([4, 5] ∙ đ?‘Ľ + 1) â „2 [4,5] đ?‘Ľâ†’0 2 = lim

[4,5]

đ?‘Ľâ†’0 2√([4, 5]

∙ đ?‘Ľ + 1) [4,5] 4 5 = = [ , ] = [2,2.5] 2 2 2

1

= đ??ź ∙ = đ??ź ∙ ∞ = ∞ ∙ đ??ź.

đ?‘Ľ

Solution: The above limit will be solved firstly by using the L'HĂ´pital's rule and secondly by using the rationalizing technique.

Note that: 0

√[4,5]∙đ?‘Ľ+1−1

Rationalizing technique [3] √[4,5] ∙ đ?‘Ľ + 1 − 1 √[4,5] ∙ 0 + 1 − 1 = đ?‘Ľâ†’0 đ?‘Ľ 0 √[4 ∙ 0, 5 ∙ 0] + 1 − 1 √[0, 0] + 1 − 1 = = 0 0 √0 + 1 − 1 0 = = 0 0 = undefined. lim

Multiply with the conjugate of the numerator:

Solution: Let � = � [2.1,2.5] → ln � = [2.1, 2.5] ln �

Huda E. Khalid, Florentin Smarandache & Ahmed K. Essa, A Neutrosophic Binomial Factorial Theorem with their Refrains

Neutrosophic Sets and Systems, Vol. 14, 2016

√[4, 5]đ?‘Ľ + 1 − 1 √[4, 5]đ?‘Ľ + 1 + 1 ∙ đ?‘Ľâ†’0 đ?‘Ľ √[4, 5]đ?‘Ľ + 1 + 1 lim

2

= lim

�→0

= lim

�→0

= lim

�→0

= lim

�→0

=

(√[4, 5]đ?‘Ľ + 1) − (1)2 đ?‘Ľ (√[4, 5]đ?‘Ľ + 1 + 1) [4, 5] ∙ đ?‘Ľ + 1 − 1 đ?‘Ľ ∙ (√[4, 5]đ?‘Ľ + 1 + 1) [4, 5] ∙ đ?‘Ľ đ?‘Ľ ∙ (√[4, 5]đ?‘Ľ + 1 + 1) [4, 5]

(√[4, 5] ∙ 0 + 1 + 1)

=

[4, 5] √1 + 1

[4, 5] 4 5 = = [ , ] = [2, 2.5].

2 2 2 Identical results.

Example (3.5) Find the value of the following neutrosophic limit lim

đ?‘Ľ 2 +3đ?‘Ľâˆ’[1,2]đ?‘Ľâˆ’[3,6] đ?‘Ľ+3

đ?‘Ľâ†’−3

using more than one

technique .

Analytical technique [1], [3] lim

đ?‘Ľâ†’−3

Again, Solving by using L'HĂ´pital's rule đ?‘Ľ 2 + 3đ?‘Ľ − [1, 2]đ?‘Ľ − [3, 6] đ?‘Ľâ†’−3 đ?‘Ľ+3 2 đ?‘Ľ + 3 − [1, 2] = lim đ?‘Ľâ†’−3 1 2 (−3) + 3 − [1, 2] = lim đ?‘Ľâ†’−3 1 = −6 + 3 − [1, 2] = −3 − [1, 2] = [−3 − 1, −3 − 2] = [−5, −4] lim

The above two methods are identical in results.

(√[4, 5]đ?‘Ľ + 1 + 1) [4, 5]

9

đ?‘Ľ 2 +3đ?‘Ľâˆ’[1,2]đ?‘Ľâˆ’[3,6] đ?‘Ľ+3

By substituting đ?‘Ľ= -3 , (−3)2 + 3 ∙ (−3) − [1, 2] ∙ (−3) − [3, 6] lim đ?‘Ľâ†’−3 −3 + 3 9 − 9 − [1 ∙ (−3), 2 ∙ (−3)] − [3, 6] = 0 0 − [−6, −3] − [3, 6] [3, 6] − [3,6] = = 0 0 [3 − 6, 6 − 3] [−3, 3] = = , 0 0 0 which has undefined operation , since 0 ∈ 0

[−3, 3]. Then we factor out the numerator, and simplify: đ?‘Ľ 2 + 3đ?‘Ľ − [1, 2]đ?‘Ľ − [3, 6] lim = đ?‘Ľâ†’−3 đ?‘Ľ+3 (đ?‘Ľ − [1, 2]) ∙ (đ?‘Ľ + 3) lim = lim (đ?‘Ľ − [1,2]) đ?‘Ľâ†’−3 đ?‘Ľâ†’−3 (đ?‘Ľ + 3) = −3 − [1,2] = [−3, −3] − [1,2] = −([3,3] + [1,2]) = [−5, −4].

4 New Theorems in Neutrosophic Limits Theorem (4.1) (Binomial Factorial ) 1

lim (đ??ź + )đ?‘Ľ = đ??źđ?‘’ ; I is the literal indeterminacy, đ?‘Ľ

đ?‘Ľâ†’∞

e = 2.7182828 Proof 1 đ?‘Ľ 1 0 1 1 đ?‘Ľ đ?‘Ľ (đ??ź + ) = ( ) đ??ź đ?‘‹ ( ) + ( ) đ??ź đ?‘‹âˆ’1 ( ) 0 1 đ?‘Ľ đ?‘Ľ đ?‘Ľ 2 3 1 1 đ?‘Ľ đ?‘Ľ + ( ) đ??ź đ?‘‹âˆ’2 ( ) + ( ) đ??ź đ?‘‹âˆ’3 ( ) 3 2 đ?‘Ľ đ?‘Ľ 1 4 đ?‘Ľ + ( ) đ??ź đ?‘‹âˆ’4 ( ) + â‹Ż 4 đ?‘Ľ 1 đ??ź 1 = đ??ź + đ?‘Ľ. đ??ź. + (1 − ) đ?‘Ľ 2! đ?‘Ľ đ??ź 1 2 đ??ź 1 2 + (1 − ) (1 − ) + (1 − ) (1 − ) 3! đ?‘Ľ đ?‘Ľ 4! đ?‘Ľ đ?‘Ľ 3 (1 − ) + â‹Ż đ?‘Ľ 1 It is clear that → 0 đ?‘Žđ?‘ đ?‘Ľ → ∞ đ?‘Ľ

1

đ??ź

đ?‘Ľ

2!

∴ lim(đ??ź − )đ?‘Ľ = đ??ź + đ??ź + đ?‘Ľâ†’∞

∑∞ đ?‘›=1

+

đ??ź 3!

+

đ??ź 4!

+â‹Ż=đ??ź+

đ??źđ?‘› đ?‘›!

1

1

∴ lim(đ??ź + )đ?‘Ľ = đ??źđ?‘’, where e = 1 + ∑ ∞ đ?‘›=1 đ?‘›! , I is the đ?‘Ľ đ?‘Ľâ†’∞

literal indeterminacy.

Corollary (4.1.1) 1

lim(đ??ź + đ?‘Ľ)đ?‘Ľ = đ??źđ?‘’

�→0

Proof:Put đ?‘Ś =

1 đ?‘Ľ

It is obvious that đ?‘Ś → ∞ , as đ?‘Ľ → 0 1

1

∴ lim(đ??ź + đ?‘Ľ)đ?‘Ľ = lim (đ??ź + )đ?‘Ś = đ??źđ?‘’ đ?‘Ľâ†’0

đ?‘Śâ†’∞

đ?‘Ś

( using Th. 4.1 ) Corollary (4.1.2) đ?‘˜

lim (đ??ź + )đ?‘Ľ = đ??źđ?‘’đ?‘˜ , where k > 0 & đ?‘˜ ≠0 , I is the đ?‘Ľ

đ?‘Ľâ†’∞

literal indeterminacy.

Huda E. Khalid, Florentin Smarandache & Ahmed K. Essa, A Neutrosophic Binomial Factorial Theorem with their Refrains

Neutrosophic Sets and Systems, Vol. 14, 2016

10

𝑦 1 = 𝑙𝑛𝑎.

1 ln(𝑦 + 𝐼) ln(𝑦 + 𝐼) 𝑦 1 = 𝑙𝑛𝑎.

1 ln(𝑦 + 𝐼)𝑦 (ln 𝑎)(𝐼𝑎 𝑥 − 𝐼) 1 ∴ lim = 𝑙𝑛𝑎 1 𝑥→0 𝑥𝑙𝑛𝑎 + 𝑙𝑛𝐼 lim 𝑙𝑛(𝑦 + 𝐼)𝑦

Proof

= 𝑙𝑛𝑎.

𝑘 𝑥 𝑘 )𝑘 ]

𝑘 lim (𝐼 + )𝑥 = lim [(𝐼 + 𝑥→∞ 𝑥 𝑥 𝑘 𝑘 Put 𝑦 = → 𝑥𝑦 = 𝑘 → 𝑥 = 𝑥→∞

𝑥

𝑦

𝑦 → 0 𝑎𝑠 𝑥 → ∞

Note that

1 𝑘

𝑘 𝑥

∴ lim (𝐼 + ) = lim [(𝐼 + 𝑦)𝑦 ] 𝑥

𝑥→∞

𝑦→0

𝑦→0

(using corollary 4.1.1 ).

1 𝑘 𝑦

𝑘

𝑘 𝑘

= [lim (𝐼 + 𝑦) ] = (𝐼𝑒) = 𝐼 𝑒 = 𝐼𝑒

𝑦→0

Corollary (4.1.3) 1 𝑘

𝑘

lim(𝐼 + )𝑥 = (𝐼𝑒) = √𝐼𝑒 ,

𝑥→0 𝑘

where 𝑘 ≠ 1 & 𝑘 > 0.

Proof The immediate substitution of the value of 𝑥 in the above limit gives indeterminate form 𝐼 ∞ , 𝑥 1

0 1

i.e. lim(𝐼 + )𝑥 = lim(𝐼 + )0 = 𝐼∞ 𝑘

𝑥→0

1

= 𝑙𝑛𝑎

𝑥 1

𝑘

𝑥→0

So we need to treat this value as follow:1

1

=

𝑙𝑛(𝐼𝑒)

using corollary (4.1.1)

𝑙𝑛𝑎 𝑙𝑛𝑎 = 𝑙𝑛 𝐼 + 𝑙𝑛𝑒 𝑙𝑛𝐼 + 1

Corollary (4.2.1) 𝐼𝑎𝑘𝑥 − 𝐼 𝑘 𝑙𝑛𝑎 lim = 𝑙𝑛𝐼 𝑥→0 1 + 𝑙𝑛𝐼 𝑥+ 𝑙𝑛𝑎𝑘 Proof 𝑦 Put 𝑦 = 𝑘𝑥 → 𝑥 = 𝑘

𝑥 1 𝑥 𝑘 𝑘 𝑥 𝑘 𝑘 lim(𝐼 + )𝑥 = lim [(𝐼 + )𝑥 ] = [lim (𝐼 + )𝑥 ] 𝑥→0 𝑥→0 𝑥→0 𝑘 𝑘 𝑘 𝑥 1 1 put 𝑦 = → 𝑥 = 𝑘𝑦 → =

𝑦 → 0 𝑎𝑠 𝑥 → 0

As 𝑥 → 0 , 𝑦 → 0

using Th. (4.2) 𝑙𝑛𝑎 = 𝑘. ( ) 1 + 𝑙𝑛𝐼

𝑘

𝑥

1

𝑥 𝑥 lim (𝐼 + ) = lim [(𝐼 + 𝑥→0 𝑦→0 𝑘

𝑘𝑦

1 1 𝑘 𝑦)𝑦 ]

= [lim (𝐼 + 𝑦→0

1 1 𝑘 𝑦)𝑦 ]

Using corollary (4.1.1) 𝐼

𝑘

= (𝐼𝑒)𝑘 = √𝐼𝑒 Theorem (4.2) lim

1

𝑙𝑛 lim (𝑦 + 𝐼)𝑦

𝑦→0

𝑥→0

1

= 𝑙𝑛𝑎 .

𝑘

(𝑙𝑛𝑎)[𝐼𝑎𝑥 −𝐼] 𝑥𝑙𝑛𝑎+𝑙𝑛𝐼

Where Note that

=

𝑙𝑛𝑎 1+𝑙𝑛𝐼

𝑎 > 0, 𝑎 ≠ 1 lim

𝑥→0

(𝑙𝑛𝑎)[𝐼𝑎𝑥 −𝐼] 𝑥𝑙𝑛𝑎+𝑙𝑛𝐼

= lim

𝐼𝑎𝑥 −𝐼

𝑥→0 𝑥+

𝑙𝑛𝐼 𝑙𝑛𝑎

Proof Let 𝑦 = 𝐼𝑎 𝑥 − 𝐼 → 𝑦 + 𝐼 = 𝐼𝑎 𝑥 → ln(𝑦 + 𝐼) = ln 𝐼 + ln 𝑎 𝑥 → ln(𝑦 + 𝐼) = ln 𝐼 + 𝑥𝑙𝑛𝑎 → ln(𝑦 + 𝐼) − 𝑙𝑛𝐼 𝑥= 𝑙𝑛𝑎 (ln 𝑎)(𝐼𝑎 𝑥 − 𝐼) (𝐼𝑎𝑥 − 𝐼) = 𝑙𝑛𝐼 𝑥𝑙𝑛𝑎 + 𝑙𝑛𝐼 𝑥+ 𝑙𝑛𝑎 𝑦 = ln(𝑦 + 𝐼) − 𝑙𝑛𝐼 𝑙𝑛𝐼 + 𝑙𝑛𝑎 𝑙𝑛𝑎

lim

𝐼𝑎𝑘𝑥 −𝐼 𝑙𝑛𝐼 𝑙𝑛𝑎𝑘

𝑥→0 𝑥+

𝐼𝑎𝑦 −𝐼

= lim 𝑦

𝑙𝑛𝐼 𝑘 𝑘 𝑙𝑛𝑎

𝑦→0 +

= 𝑘. lim

𝐼𝑎𝑦 −𝐼

𝑦→0 𝑦+

𝑙𝑛𝐼 𝑙𝑛𝑎

Corollary (4.2.2) 𝐼𝑒 𝑥 − 𝐼 1 lim = 𝑥→0 𝑥 + 𝑙𝑛𝐼 1 + 𝑙𝑛𝐼 Proof Let 𝑦 = 𝐼𝑒 𝑥 − 𝐼 , 𝑦 → 0 𝑎𝑠 𝑥 → 0 𝑦 + 𝐼 = 𝐼𝑒 𝑥 → ln(𝑦 + 𝐼) = 𝑙𝑛𝐼 + 𝑥 𝑙𝑛𝑒 𝑥 = ln(𝑦 + 𝐼) − 𝑙𝑛𝐼 𝐼𝑒 𝑥 − 𝐼 𝑦 ∴ = 𝑥 + 𝑙𝑛𝐼 ln(𝑦 + 𝐼) − 𝑙𝑛𝐼 + 𝑙𝑛𝐼 1 = 1 ln(𝑦 + 𝐼) 𝑦 1 = 1 ln(𝑦 + 𝐼)𝑦 𝐼𝑒 𝑥 − 𝐼 1 ∴ lim = lim 1 𝑥→0 𝑥 + 𝑙𝑛𝐼 𝑦→0 ln(𝑦 + 𝐼)𝑦 1 = 1 ln lim (𝑦 + 𝐼)𝑦 𝑦→0

using corollary (4.1.1) 1 1 1 = = ln(𝐼𝑒) 𝑙𝑛𝐼 + 𝑙𝑛𝑒 𝑙𝑛𝐼 + 1

Huda E. Khalid, Florentin Smarandache & Ahmed K. Essa, A Neutrosophic Binomial Factorial Theorem with their Refrains

11

Neutrosophic Sets and Systems, Vol. 14, 2016

Corollary (4.2.3) đ??źđ?‘’ đ?‘˜đ?‘Ľ − đ??ź đ?‘˜ lim = đ?‘™đ?‘›đ??ź đ?‘Ľâ†’0 1 + đ?‘™đ?‘›đ??ź đ?‘Ľ+ đ?‘˜ Proof đ?‘Ś let đ?‘Ś = đ?‘˜đ?‘Ľ → đ?‘Ľ =

5 Numerical Examples

đ?‘Ś → 0 đ?‘Žđ?‘ đ?‘Ľ → 0

�→0 �+

Example (5.1) Evaluate the limit lim

đ??źđ?‘’ đ?‘˜đ?‘Ľ −đ??ź

đ?‘™đ?‘›đ??ź đ?‘Ľâ†’0 đ?‘Ľ+ đ?‘˜

đ??źđ?‘’ đ?‘Ś −đ??ź

= lim đ?‘Ś

đ?‘™đ?‘›đ??ź đ?‘Śâ†’0 + đ?‘˜ đ?‘˜

Solution lim

= đ?‘˜. lim

đ??źđ?‘’ đ?‘Ś −đ??ź

đ?‘Śâ†’0 đ?‘Ś+đ?‘™đ?‘›đ??ź

using

Corollary (4.2.2) to get 1 đ?‘˜ = đ?‘˜. ( )= 1 + đ?‘™đ?‘›đ??ź 1 + đ?‘™đ?‘›đ??ź Theorem (4.3) ln(đ??ź + đ?‘˜đ?‘Ľ) lim = đ?‘˜(1 + đ?‘™đ?‘›đ??ź) đ?‘Ľâ†’0 đ?‘Ľ Proof ln(đ??ź + đ?‘˜đ?‘Ľ) ln(đ??ź + đ?‘˜đ?‘Ľ) − đ?‘™đ?‘›đ??ź + đ?‘™đ?‘›đ??ź lim = lim đ?‘Ľâ†’0 đ?‘Ľâ†’0 đ?‘Ľ đ?‘Ľ Let đ?‘Ś = ln(đ??ź + đ?‘˜đ?‘Ľ) − đ?‘™đ?‘›đ??ź → đ?‘Ś + đ?‘™đ?‘›đ??ź = ln(đ??ź + đ?‘˜đ?‘Ľ) đ?‘’ đ?‘Ś đ?‘’ đ?‘™đ?‘›đ??ź − đ??ź đ??ź đ?‘’ đ?‘Ś − đ??ź đ?‘’ đ?‘Ś+đ?‘™đ?‘›đ??ź = đ??ź + đ?‘˜đ?‘Ľ → đ?‘Ľ = = đ?‘˜ đ?‘˜ đ?‘Ś → 0 đ?‘Žđ?‘ đ?‘Ľ → 0 ln(đ??ź + đ?‘˜đ?‘Ľ) − đ?‘™đ?‘›đ??ź + đ?‘™đ?‘›đ??ź lim đ?‘Ľâ†’0 đ?‘Ľ đ?‘Ś + đ?‘™đ?‘›đ??ź = lim đ?‘Ś đ?‘Śâ†’0 đ??ź đ?‘’ − đ??ź đ?‘˜ đ?‘˜ đ?‘˜ lim đ??ź đ?‘’đ?‘Śâˆ’đ??ź = đ??ź đ?‘’đ?‘Śâˆ’đ??ź đ?‘Śâ†’0

lim

đ?‘Śâ†’0( đ?‘Ś+đ?‘™đ?‘›đ??ź

đ?‘Ś+đ?‘™đ?‘›đ??ź

)

using corollary (4.2.2) to get the result đ?‘˜ = = đ?‘˜(1 + đ?‘™đ?‘›đ??ź) 1 1 + đ?‘™đ?‘›đ??ź Theorem (4.4) Prove that, for any two real numbers đ?‘Ž, đ?‘? lim

đ??źađ?‘Ľ −đ??ź

đ?‘Ľâ†’0 đ??źbđ?‘Ľâˆ’đ??ź

limit conclude that

0 0

,so we need to treat it as

follow: đ?‘™đ?‘›a[đ??źađ?‘Ľ − đ??ź] đ?‘Ľđ?‘™đ?‘›a + đ?‘™đ?‘›đ??ź ∗ đ??źađ?‘Ľ − đ??ź đ?‘™đ?‘›đ??ź đ?‘™đ?‘›a lim đ?‘Ľ = lim đ?‘Ľđ?‘™đ?‘›a + đ?‘Ľ đ?‘Ľâ†’0 đ??źb − đ??ź đ?‘Ľâ†’0 đ?‘™đ?‘›b[đ??źb − đ??ź] đ?‘Ľđ?‘™đ?‘›b + đ?‘™đ?‘›đ??ź ∗ đ?‘Ľđ?‘™đ?‘›b + đ?‘™đ?‘›đ??ź đ?‘™đ?‘›b đ?‘Ľ đ?‘™đ?‘›a[đ??źa − đ??ź] lim lim ( đ?‘Ľđ?‘™đ?‘›a + đ?‘™đ?‘›đ??ź) đ?‘™đ?‘›b đ?‘Ľđ?‘™đ?‘›a + đ?‘™đ?‘›đ??ź đ?‘Ľâ†’0 = đ?‘Ľâ†’đ?‘Ľ ∗ ∗ đ?‘™đ?‘›b[đ??źb đ?‘Ľ − đ??ź] lim( đ?‘Ľđ?‘™đ?‘›b + đ?‘™đ?‘›đ??ź) đ?‘™đ?‘›a lim đ?‘Ľâ†’0 đ?‘Ľâ†’đ?‘Ľ đ?‘Ľđ?‘™đ?‘›b + đ?‘™đ?‘›đ??ź (using Th.(4.2) twice (first in numerator second in denominator )) đ?‘™đ?‘›a 1+đ?‘™đ?‘›đ??ź đ?‘™đ?‘›b 1+đ?‘™đ?‘›đ??ź

∗

đ??ź54đ?‘Ľ −đ??ź đ?‘™đ?‘›đ??ź đ?‘™đ?‘›54

=

4đ?‘™đ?‘›5 1+đ?‘™đ?‘›đ??ź

(using corollary 4. 2.1)

Example (5.2) Evaluate the limit lim

đ??źđ?‘’ 4đ?‘Ľ −đ??ź

đ?‘Ľâ†’0 đ??ź32đ?‘Ľ −đ??ź

Solution đ?‘™đ?‘›3[đ??źđ?‘’ 4đ?‘Ľ − đ??ź] đ?‘™đ?‘›đ??ź ∗ (đ?‘Ľ + ) 4 đ?‘™đ?‘›đ??ź 4đ?‘Ľ (đ?‘Ľ + ) đ??źđ?‘’ − đ??ź 4 lim = lim đ?‘Ľâ†’0 đ??ź32đ?‘Ľ − đ??ź đ?‘Ľâ†’0 đ?‘™đ?‘›3[đ??ź32đ?‘Ľ − đ??ź] đ?‘™đ?‘›đ??ź ∗ (đ?‘Ľ + ) đ?‘™đ?‘›đ??ź đ?‘™đ?‘›32 (đ?‘Ľ + ) 2 đ?‘™đ?‘›3 đ?‘™đ?‘›3[đ??źđ?‘’ 4đ?‘Ľ − đ??ź] lim đ?‘™đ?‘›đ??ź đ?‘™đ?‘›đ??ź đ?‘Ľâ†’0 lim (đ?‘Ľ + ) (đ?‘Ľ + ) 4 4 đ?‘Ľâ†’0 = ∗ 2đ?‘Ľ đ?‘™đ?‘›đ??ź đ?‘™đ?‘›3[đ??ź3 − đ??ź] lim (đ?‘Ľ + ) lim đ?‘™đ?‘›32 đ?‘™đ?‘›đ??ź đ?‘Ľâ†’0 đ?‘Ľâ†’0 (đ?‘Ľ + ) đ?‘™đ?‘›32 (using corollary (4.2.3) on numerator & corollary (4.2.1) on denominator ) 4 đ?‘™đ?‘›đ??ź 1 + đ?‘™đ?‘›đ??ź = ∗ 4 = 1.

2đ?‘™đ?‘›3 đ?‘™đ?‘›đ??ź 1 + đ?‘™đ?‘›đ??ź đ?‘™đ?‘›32 5 Conclusion In this article, we introduced for the first time a new version of binomial factorial theorem containing the literal indeterminacy (I). This theorem enhances three corollaries. As a conjecture for indeterminate forms in classical calculus, ten of new indeterminate forms in Neutrosophic calculus had been constructed.

Finally, various examples had been solved.

= 1 , where đ?‘Ž, đ?‘? > 0 & đ?‘Ž, đ?‘? ≠1

Proof The direct substitution of the value đ?‘Ľ in the above

=

đ?‘™đ?‘›đ??ź đ?‘™đ?‘›54

�→0 �+

đ?‘˜

lim

đ??ź54đ?‘Ľ −đ??ź

đ?‘™đ?‘›đ??ź đ?‘™đ?‘›đ??ź

∗

��b ��a

= 1.

References [1] F. Smarandache. Neutrosophic Precalculus and Neutrosophic Calculus. EuropaNova Brussels, 2015.

[2] F. Smarandache. Introduction to Neutrosophic Statistics. Sitech and Education Publisher, Craiova, 2014.

[3] H. E. Khalid & A. K. Essa. Neutrosophic Precalculus and Neutrosophic Calculus. Arabic version of the book. Pons asbl 5, Quai du Batelage, Brussells, Belgium, European Union 2016.

[4] H. Anton, I. Bivens & S. Davis, Calculus, 7th Edition, John Wiley & Sons, Inc. 2002.

Received: November 7, 2016. Accepted: November 14, 2016

Huda E. Khalid, Florentin Smarandache & Ahmed K. Essa, A Neutrosophic Binomial Factorial Theorem with their Refrains