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

Essa, Ahmed K.

Khalid, Huda E.

Smarandache, Florentin

PhilPapers

Neutrosophic Sets and Systems, Vol. 14, 2016Huda E. Khalid, Florentin Smarandache & Ahmed K. Essa, A Neutrosophic Binomial Factorial Theorem with their Refrains University of New MexicoA Neutrosophic Binomial Factorial Theorem with their Refrains  Huda E. Khalid1       Florentin Smarandache2       Ahmed K. Essa3 1 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) Q 1 What are the types of indeterminacy? There exist two types of indeterminacy a. Literal indeterminacy (I).As example:   2 + 3𝐼  (1) b. Numerical indeterminacy.As example: 𝑥(0.6,0.3,0.4) ∈ 𝐴, (2) meaning that the indeterminacy membership = 0.3.         Other examples for the indeterminacy com-ponent can be seen in functions: 𝑓(0) = 7 𝑜𝑟 9  or  𝑓(0  𝑜𝑟  1) = 5  or 𝑓(𝑥) = [0.2, 0.3] 𝑥2 … etc. Q 2 What is the values of 𝐼 to the rational power? 1. Let√𝐼 = 𝑥 + 𝑦 𝐼0 + 𝐼 = 𝑥2 + (2𝑥𝑦 + 𝑦2)𝐼  𝑥 = 0, 𝑦 = ±1. (3) In general, √𝐼2𝑘= ±𝐼 (4) where 𝑘 ∈ 𝑧+ = {1,2,3, … }. 2. Let√𝐼3= 𝑥 + 𝑦 𝐼 0 + 𝐼 = 𝑥3 + 3𝑥2𝑦 𝐼 + 3𝑥𝑦2 𝐼2 + 𝑦3𝐼3 0 + 𝐼 = 𝑥3 + (3𝑥2𝑦 + 3𝑥𝑦2  + 𝑦3)𝐼  𝑥 = 0, 𝑦 = 1 →  √𝐼3= 𝐼. (5) In general, √𝐼2𝑘+1= 𝐼, (6) where 𝑘 ∈ 𝑧+ = {1,2,3, … }. Basic Notes  1. A component I to the zero power isundefined value, (i.e. 𝐼0 is undefined),since 𝐼0 = 𝐼1+(−1) = 𝐼1 ∗ 𝐼−1 =𝐼𝐼 which isimpossible case (avoid to divide by 𝐼). 2. The value of 𝐼 to the negative power isundefined value (i.e. 𝐼−𝑛  , 𝑛 > 0 isundefined).Q 3 What are the indeterminacy forms in neutros-ophic calculus? In classical calculus, the indeterminate forms are [4]: 00,∞∞, 0 ∙ ∞ , ∞0, 00, 1∞, ∞ − ∞. (7) 7Huda E. Khalid, Florentin Smarandache & Ahmed K. Essa, A Neutrosophic Binomial Factorial Theorem with their Refrains The form 0 to the power 𝐼  (i.e. 0𝐼 ) is an indeterminate form in Neutrosophic calculus; it is tempting to argue that an indeterminate form of type 0𝐼 has zero value since "zero to any power is zero". However, this is fallacious, since 0𝐼 is not a power of number, but rather a statement about limits. Q 4 What about the form 1𝐼? The base "one" pushes the form 1𝐼 to one while the power 𝐼 pushes the form 1𝐼 to I, so 1𝐼 is an indeterminate form in neutrosophic calculus. Indeed, the form 𝑎𝐼, 𝑎 ∈ 𝑅 is always an indeterminate form. Q 5 What is the value of 𝑎𝐼  , 𝑤ℎ𝑒𝑟𝑒 𝑎 ∈ 𝑅? Let 𝑦1 = 2𝑥 , 𝑥 ∈ 𝑅 , 𝑦2 = 2𝐼; it is obvious thatlim𝑥→∞2𝑥 = ∞  , lim𝑥→−∞2𝑥 = 0  , lim𝑥→02𝑥 = 1; while we cannot determine if 2𝐼 → ∞ 𝑜𝑟 0 𝑜𝑟 1, therefore we can say that  𝑦2 = 2𝐼 indeterminateform in Neutrosophic calculus. The same for 𝑎𝐼 ,where 𝑎 ∈ 𝑅 [2]. 2 Indeterminate forms in Neutrosophic Logic It is obvious that there are seven types of indeterminate forms in classical calculus [4], 00,∞∞, 0. ∞, 00, ∞0, 1∞, ∞ − ∞.  As a conjecture, we can say that there are ten forms of the indeterminate forms in Neutrosophic calculus  𝐼0 , 0𝐼 ,𝐼0, 𝐼 ∙ ∞,∞𝐼 , ∞𝐼 , 𝐼∞, 𝐼𝐼 , 𝑎𝐼(𝑎 ∈ 𝑅), ∞ ± 𝑎 ∙ 𝐼  . Note that:   𝐼0= 𝐼 ∙10= 𝐼 ∙ ∞ = ∞ ∙ 𝐼. 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 lim𝑥→0𝑓(𝑥), where 𝑓(𝑥) = 𝑥[2.1,2.5].Solution: Let 𝑦 = 𝑥[2.1,2.5]   → ln 𝑦 = [2.1, 2.5] ln 𝑥∴ lim𝑥→0ln 𝑦 = lim𝑥→0[2.1, 2.5]1ln 𝑥= [2.1, 2.5]1ln 0=[2.1, 2.5]1−∞=[2.1, 2.5]−0= [2.1−0,2.5−0] = (−∞, −∞)= −∞ Hence 𝑦 = 𝑒−∞ = 0 OR it can be solved briefly by 𝑦 = 𝑥[2.1,2.5] = [02.1, 02.5] = [0,0] = 0. Example (3.2) lim𝑥→[9,11][3.5,5.9]𝑥[1,2] =  [3.5,5.9] [9,11][1,2] =[3.5,5.9] [91, 112] =  [(3.5)(9), (5.9)(121)] = [31.5,713.9]. Example (3.3) lim𝑥→∞[3.5,5.9] 𝑥[1,2] = [3.5,5.9]  ∞[1,2]= [3.5,5.9] [∞1, ∞2]=  [3.5 ∙ (∞) ,5.9 ∙ (∞)]= (∞, ∞) = ∞. Example (3.4) Find the following limit using more than one technique lim𝑥→0√[4,5]∙𝑥+1−1𝑥 . Solution:  The above limit will be solved firstly by using the L'Hôpital's rule and secondly by using the rationalizing  technique. Using L'Hôpital's rule lim𝑥→012([4, 5] ∙ 𝑥 + 1)−12⁄  [4,5]= lim𝑥→0 [4,5]2√([4, 5] ∙ 𝑥 + 1)= [4,5]2=  [42,52] =  [2,2.5] Rationalizing technique [3] lim𝑥→0√[4,5] ∙ 𝑥 + 1 − 1𝑥=√[4,5] ∙ 0 + 1 − 10=√[4 ∙ 0, 5 ∙ 0] + 1 − 10=√[0, 0] + 1 − 10=√0 + 1 − 10=00= undefined. Multiply with the conjugate of the numerator: 8 Neutrosophic Sets and Systems, Vol. 14, 2016lim𝑥→0√[4, 5]𝑥 + 1 − 1𝑥∙√[4, 5]𝑥 + 1 + 1√[4, 5]𝑥 + 1 + 1= lim𝑥→0(√[4, 5]𝑥 + 1)2− (1)2𝑥 (√[4, 5]𝑥 + 1 + 1)= lim𝑥→0[4, 5] ∙ 𝑥 + 1 − 1𝑥 ∙ (√[4, 5]𝑥 + 1 + 1)= lim𝑥→0[4, 5] ∙ 𝑥𝑥 ∙ (√[4, 5]𝑥 + 1 + 1)= lim𝑥→0[4, 5](√[4, 5]𝑥 + 1 + 1)=[4, 5](√[4, 5] ∙ 0 + 1 + 1)=[4, 5]√1 + 1=[4, 5]2= [42,52] = [2, 2.5]. Identical results. Example (3.5) Find the value of  the following neutrosophic limit    lim𝑥→−3𝑥2+3𝑥−[1,2]𝑥−[3,6]𝑥+3using more than onetechnique . Analytical technique [1], [3] lim𝑥→−3𝑥2+3𝑥−[1,2]𝑥−[3,6]𝑥+3By  substituting  𝑥= -3 , lim𝑥→−3(−3)2 + 3 ∙ (−3) − [1, 2] ∙ (−3) − [3, 6]−3 + 3=9 − 9 − [1 ∙ (−3), 2 ∙ (−3)] − [3, 6]0=0 − [−6, −3] − [3, 6]0=[3, 6] − [3,6]0=[3 − 6, 6 − 3]0=[−3, 3]0, which has  undefined operation00, since 0 ∈[−3, 3]. Then we factor out the numerator, and simplify: lim𝑥→−3𝑥2 + 3𝑥 − [1, 2]𝑥 − [3, 6]𝑥 + 3= lim𝑥→−3(𝑥 − [1, 2]) ∙ (𝑥 + 3)(𝑥 + 3)= lim𝑥→−3(𝑥 − [1,2]) = −3 − [1,2] = [−3, −3] − [1,2] =  −([3,3] + [1,2]) = [−5, −4]. Again, Solving by using L'Hôpital's rule lim𝑥→−3𝑥2 + 3𝑥 − [1, 2]𝑥 − [3, 6]𝑥 + 3= lim𝑥→−32 𝑥 + 3 − [1, 2]1= lim𝑥→−32 (−3) + 3 − [1, 2]1= −6 + 3 − [1, 2]= −3 − [1, 2]= [−3 − 1, −3 − 2]= [−5, −4] The above two methods are identical in results. 4 New Theorems in Neutrosophic Limits Theorem (4.1) (Binomial  Factorial ) lim𝑥→∞(𝐼 +1𝑥)𝑥 = 𝐼𝑒  ;  I is the literal indeterminacy, e = 2.7182828 Proof (𝐼 +1𝑥)𝑥= (𝑥0) 𝐼𝑋 (1𝑥)0+ (𝑥1) 𝐼𝑋−1 (1𝑥)1+ (𝑥2) 𝐼𝑋−2 (1𝑥)2+ (𝑥3) 𝐼𝑋−3 (1𝑥)3+ (𝑥4) 𝐼𝑋−4 (1𝑥)4+ ⋯ = 𝐼 + 𝑥. 𝐼.1𝑥+𝐼2!(1 −1𝑥) +𝐼3!(1 −1𝑥) (1 −2𝑥) +𝐼4!(1 −1𝑥) (1 −2𝑥) (1 −3𝑥) + ⋯ It is clear that   1𝑥→ 0  𝑎𝑠  𝑥  → ∞ ∴ lim𝑥→∞(𝐼 −1𝑥)𝑥 = 𝐼 + 𝐼 +𝐼2!+𝐼3!+𝐼4!+ ⋯ = 𝐼 +∑𝐼𝑛𝑛!∞𝑛=1∴ lim𝑥→∞(𝐼 +1𝑥)𝑥 = 𝐼𝑒, where e = 1 + ∑ 1𝑛!∞𝑛=1  , I is theliteral indeterminacy. Corollary (4.1.1) lim𝑥→0(𝐼 + 𝑥)1𝑥 = 𝐼𝑒 Proof:- Put 𝑦 =1𝑥It is obvious that   𝑦 → ∞ , as 𝑥 → 0 ∴ lim𝑥→0(𝐼 + 𝑥)1𝑥 = lim𝑦→∞(𝐼 +1𝑦)𝑦 = 𝐼𝑒  ( using Th. 4.1 ) Corollary (4.1.2) lim𝑥→∞(𝐼 +𝑘𝑥)𝑥 = 𝐼𝑒𝑘  , where k > 0 & 𝑘 ≠ 0 , I is the literal indeterminacy. Neutrosophic Sets and Systems, Vol. 14, 2016  9 Huda E. Khalid, Florentin Smarandache & Ahmed K. Essa, A Neutrosophic Binomial Factorial Theorem with their Refrains Huda E. Khalid, Florentin Smarandache & Ahmed K. Essa, A Neutrosophic Binomial Factorial Theorem with their Refrains Proof lim𝑥→∞(𝐼 +𝑘𝑥)𝑥 = lim𝑥→∞[(𝐼 +𝑘𝑥)𝑥𝑘]𝑘Put 𝑦 =𝑘𝑥→ 𝑥𝑦 = 𝑘 → 𝑥 =𝑘𝑦Note that     𝑦 → 0 𝑎𝑠 𝑥 → ∞  ∴  lim𝑥→∞(𝐼 +𝑘𝑥)𝑥= lim𝑦→0[(𝐼 + 𝑦)1𝑦]𝑘(using corollary 4.1.1 ). = [lim𝑦→0(𝐼 + 𝑦)1𝑦]𝑘= (𝐼𝑒)𝑘 = 𝐼𝑘𝑒𝑘 = 𝐼𝑒𝑘 Corollary (4.1.3) lim(𝐼 +𝑥)1𝑥 = (𝐼𝑒)1𝑘 = √𝐼𝑒𝑘  , 𝑥→0 𝑘where 𝑘 ≠ 1 & 𝑘 > 0. Proof The immediate substitution of the value of 𝑥 in the above limit gives indeterminate form 𝐼∞, i.e. lim𝑥→0(𝐼 +𝑥𝑘)1𝑥 = lim𝑥→0(𝐼 +0𝑘)10 = 𝐼∞So we need to treat this value as follow:- lim𝑥→0(𝐼 +𝑥𝑘)1𝑥 = lim𝑥→0[(𝐼 +𝑥𝑘)𝑘𝑥]1𝑘= [lim𝑥→0(𝐼 +𝑥𝑘)𝑘𝑥]1𝑘put 𝑦 =𝑥𝑘→ 𝑥 = 𝑘𝑦 →1𝑥=1𝑘𝑦As 𝑥 → 0  , 𝑦 → 0 lim𝑥→0(𝐼 +𝑥𝑘)1𝑥= lim𝑦→0[(𝐼 + 𝑦)1𝑦]1𝑘= [lim𝑦→0(𝐼 + 𝑦)1𝑦]1𝑘Using corollary (4.1.1) = (𝐼𝑒)𝐼𝑘 = √𝐼𝑒𝑘 Theorem (4.2) lim𝑥→0(𝑙𝑛𝑎)[𝐼𝑎𝑥−𝐼]𝑥𝑙𝑛𝑎+𝑙𝑛𝐼=𝑙𝑛𝑎1+𝑙𝑛𝐼Where     𝑎 > 0, 𝑎 ≠ 1 Note that         lim𝑥→0(𝑙𝑛𝑎)[𝐼𝑎𝑥−𝐼]𝑥𝑙𝑛𝑎+𝑙𝑛𝐼= lim𝑥→0𝐼𝑎𝑥−𝐼𝑥+𝑙𝑛𝐼𝑙𝑛𝑎ProofLet 𝑦 = 𝐼𝑎𝑥 − 𝐼 → 𝑦 + 𝐼 = 𝐼𝑎𝑥 → ln(𝑦 + 𝐼) = ln 𝐼 + ln 𝑎𝑥 → ln(𝑦 + 𝐼) = ln 𝐼 + 𝑥𝑙𝑛𝑎 → 𝑥 =  ln(𝑦 + 𝐼) − 𝑙𝑛𝐼𝑙𝑛𝑎(ln 𝑎)(𝐼𝑎𝑥 − 𝐼)𝑥𝑙𝑛𝑎 + 𝑙𝑛𝐼=(𝐼𝑎𝑥 − 𝐼)𝑥 +𝑙𝑛𝐼𝑙𝑛𝑎=𝑦ln(𝑦 + 𝐼) − 𝑙𝑛𝐼𝑙𝑛𝑎+𝑙𝑛𝐼𝑙𝑛𝑎= 𝑙𝑛𝑎.𝑦ln(𝑦 + 𝐼)= 𝑙𝑛𝑎.11𝑦ln(𝑦 + 𝐼)= 𝑙𝑛𝑎.1ln(𝑦 + 𝐼)1𝑦∴ lim𝑥→0 (ln 𝑎)(𝐼𝑎𝑥 − 𝐼)𝑥𝑙𝑛𝑎 + 𝑙𝑛𝐼= 𝑙𝑛𝑎 1lim𝑦→0𝑙𝑛(𝑦 + 𝐼)1𝑦= 𝑙𝑛𝑎 .1𝑙𝑛 lim𝑦→0(𝑦 + 𝐼)1𝑦= 𝑙𝑛𝑎 1𝑙𝑛(𝐼𝑒) using corollary (4.1.1) =  𝑙𝑛𝑎𝑙𝑛 𝐼 +  𝑙𝑛𝑒=𝑙𝑛𝑎𝑙𝑛𝐼 + 1Corollary (4.2.1) lim𝑥→0𝐼𝑎𝑘𝑥 − 𝐼𝑥 +𝑙𝑛𝐼𝑙𝑛𝑎𝑘=𝑘 𝑙𝑛𝑎1 + 𝑙𝑛𝐼Proof  Put 𝑦 = 𝑘𝑥 → 𝑥 = 𝑦𝑘𝑦 → 0  𝑎𝑠 𝑥 → 0 lim𝑥→0𝐼𝑎𝑘𝑥−𝐼𝑥+𝑙𝑛𝐼𝑙𝑛𝑎𝑘= lim𝑦→0𝐼𝑎𝑦−𝐼𝑦𝑘+𝑙𝑛𝐼𝑘 𝑙𝑛𝑎= 𝑘. lim𝑦→0𝐼𝑎𝑦−𝐼𝑦+𝑙𝑛𝐼 𝑙𝑛𝑎using Th. (4.2) = 𝑘. (𝑙𝑛𝑎1 + 𝑙𝑛𝐼) Corollary (4.2.2) lim𝑥→0𝐼𝑒𝑥 − 𝐼𝑥 + 𝑙𝑛𝐼=11 + 𝑙𝑛𝐼Proof Let 𝑦 =  𝐼𝑒𝑥 − 𝐼   , 𝑦 → 0  𝑎𝑠 𝑥 → 0 𝑦 + 𝐼 =  𝐼𝑒𝑥 → ln(𝑦 + 𝐼) = 𝑙𝑛𝐼 + 𝑥 𝑙𝑛𝑒 𝑥 = ln(𝑦 + 𝐼) − 𝑙𝑛𝐼 ∴  𝐼𝑒𝑥 − 𝐼 𝑥 +  𝑙𝑛𝐼=𝑦ln(𝑦 + 𝐼) − 𝑙𝑛𝐼 + 𝑙𝑛𝐼=11𝑦ln(𝑦 + 𝐼)=1ln(𝑦 + 𝐼)1𝑦∴ lim𝑥→0𝐼𝑒𝑥 − 𝐼𝑥 + 𝑙𝑛𝐼= lim𝑦→01ln(𝑦 + 𝐼)1𝑦=1ln lim𝑦→0(𝑦 + 𝐼)1𝑦using corollary (4.1.1) 1ln (𝐼𝑒)=1𝑙𝑛𝐼 + 𝑙𝑛𝑒=1𝑙𝑛𝐼 + 110 Neutrosophic Sets and Systems, Vol. 14, 2016Corollary (4.2.3) lim𝑥→0𝐼𝑒𝑘𝑥 − 𝐼𝑥 +𝑙𝑛𝐼𝑘=𝑘 1 + 𝑙𝑛𝐼Proof  let 𝑦 = 𝑘𝑥 → 𝑥 =𝑦𝑘𝑦 → 0  𝑎𝑠 𝑥 → 0 lim𝑥→0𝐼𝑒𝑘𝑥−𝐼𝑥+𝑙𝑛𝐼𝑘= lim𝑦→0𝐼𝑒𝑦−𝐼𝑦𝑘+𝑙𝑛𝐼𝑘 = 𝑘. lim𝑦→0𝐼𝑒𝑦−𝐼𝑦+𝑙𝑛𝐼   using Corollary (4.2.2) to  get = 𝑘. (11 + 𝑙𝑛𝐼) =𝑘1 + 𝑙𝑛𝐼Theorem (4.3) lim𝑥→0ln (𝐼 + 𝑘𝑥)𝑥= 𝑘(1 + 𝑙𝑛𝐼) Proof lim𝑥→0ln (𝐼 + 𝑘𝑥)𝑥= lim𝑥→0ln(𝐼 + 𝑘𝑥) − 𝑙𝑛𝐼 + 𝑙𝑛𝐼𝑥Let 𝑦 = ln(𝐼 + 𝑘𝑥) − 𝑙𝑛𝐼 → 𝑦 + 𝑙𝑛𝐼 = ln(𝐼 +𝑘𝑥) 𝑒𝑦+𝑙𝑛𝐼 = 𝐼 + 𝑘𝑥 → 𝑥 =𝑒𝑦𝑒𝑙𝑛𝐼 − 𝐼𝑘=𝐼 𝑒𝑦 − 𝐼𝑘𝑦 → 0  𝑎𝑠 𝑥 → 0 lim𝑥→0ln(𝐼 + 𝑘𝑥) − 𝑙𝑛𝐼 + 𝑙𝑛𝐼𝑥= lim𝑦→0𝑦 + 𝑙𝑛𝐼𝐼 𝑒𝑦 − 𝐼𝑘 lim𝑦→0𝑘𝑦+𝑙𝑛𝐼𝐼 𝑒𝑦−𝐼=𝑘lim𝑦→0()using corollary (4.2.2)  to get the result   =𝑘11 + 𝑙𝑛𝐼= 𝑘(1 + 𝑙𝑛𝐼) Theorem  (4.4) Prove that, for any two real numbers 𝑎, 𝑏 lim𝑥→0𝐼a𝑥−𝐼𝐼b𝑥−𝐼= 1 , where 𝑎, 𝑏 > 0 & 𝑎, 𝑏 ≠ 1 Proof The direct substitution of the value 𝑥 in the above limit conclude that  00 ,so we need to treat it as follow: lim𝑥→0𝐼a𝑥 − 𝐼𝐼b𝑥 − 𝐼= lim𝑥→0𝑙𝑛a[𝐼a𝑥 − 𝐼]𝑥𝑙𝑛a + 𝑙𝑛𝐼∗𝑥𝑙𝑛a + 𝑙𝑛𝐼𝑙𝑛a𝑙𝑛b[𝐼b𝑥 − 𝐼]𝑥𝑙𝑛b + 𝑙𝑛𝐼∗𝑥𝑙𝑛b + 𝑙𝑛𝐼𝑙𝑛b=lim𝑥→𝑥𝑙𝑛a[𝐼a𝑥 − 𝐼]𝑥𝑙𝑛a + 𝑙𝑛𝐼lim𝑥→𝑥𝑙𝑛b[𝐼b𝑥 − 𝐼]𝑥𝑙𝑛b + 𝑙𝑛𝐼∗lim𝑥→0( 𝑥𝑙𝑛a + 𝑙𝑛𝐼)lim (𝑥→0𝑥𝑙𝑛b + 𝑙𝑛𝐼)∗𝑙𝑛b𝑙𝑛a(using Th.(4.2) twice (first in numerator second in denominator )) =𝑙𝑛a1+𝑙𝑛𝐼𝑙𝑛b1+𝑙𝑛𝐼∗𝑙𝑛𝐼𝑙𝑛𝐼∗𝑙𝑛b𝑙𝑛a = 1. 5 Numerical Examples Example (5.1) Evaluate the limit lim𝑥→0𝐼54𝑥−𝐼𝑥+𝑙𝑛𝐼𝑙𝑛54Solution lim𝑥→0𝐼54𝑥−𝐼𝑥+𝑙𝑛𝐼𝑙𝑛54=4𝑙𝑛51+𝑙𝑛𝐼  (using corollary 4. 2.1) Example (5.2) Evaluate the limit lim𝑥→0𝐼𝑒4𝑥−𝐼𝐼32𝑥−𝐼Solution lim𝑥→0𝐼𝑒4𝑥 − 𝐼𝐼32𝑥 − 𝐼= lim𝑥→0𝑙𝑛3[𝐼𝑒4𝑥 − 𝐼](𝑥 +𝑙𝑛𝐼4)∗ (𝑥 +𝑙𝑛𝐼4)𝑙𝑛3[𝐼32𝑥 − 𝐼](𝑥 +𝑙𝑛𝐼𝑙𝑛32)∗ (𝑥 +𝑙𝑛𝐼𝑙𝑛32)=lim𝑥→0𝑙𝑛3[𝐼𝑒4𝑥 − 𝐼](𝑥 +𝑙𝑛𝐼4)lim𝑥→0𝑙𝑛3[𝐼32𝑥 − 𝐼](𝑥 +𝑙𝑛𝐼𝑙𝑛32)∗lim𝑥→0(𝑥 +𝑙𝑛𝐼4)lim𝑥→0(𝑥 +𝑙𝑛𝐼𝑙𝑛32)(using corollary (4.2.3) on numerator & corollary (4.2.1) on denominator ) =41 + 𝑙𝑛𝐼2𝑙𝑛31 + 𝑙𝑛𝐼∗𝑙𝑛𝐼4𝑙𝑛𝐼𝑙𝑛32= 1. 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. 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 Pre-calculus 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. Neutrosophic Sets and Systems, Vol. 14, 2016  11Huda E. Khalid, Florentin Smarandache & Ahmed K. Essa, A Neutrosophic Binomial Factorial Theorem with their Refrains Received: November 7, 2016.   Accepted: November 14, 2016𝑦+𝑙𝑛𝐼𝐼 𝑒𝑦−𝐼

A Neutrosophic Binomial Factorial Theorem with their Refrains

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A Neutrosophic Binomial Factorial Theorem with their Refrains

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