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Evaluating Indeterminate Limits without

Abstract: This paper proposes an algebraic method of evaluating certain indeterminate limits. This method does not involve the use of calculus. It is, however, built on calculus. Unlike methods such as L’Hospital’s rule, this proposed methodology does not require a fractional format of functions to be used.

Methodology: Step 1: Convert all trigonometric functions to sine and , or make cosine a coefficient. Step 2: Eliminate all constants in the numerator and denominator. Step 3: Take all remaining analytical functions to their (or in the case of zero, 1) multiplied by any coefficients (putting the limit back in, and if one gets zero, then one is used instead.) (Note: If there is a problem where zero is in the denominator, and there are only trigonometric 3 functions in the expression, ( lim sin (x) ), then that zero will go to one.). x→0 sin(x)−tan(x)

Step 4: Then, for functions of different degrees, one may take the exponent down and take its after dividing the exponents. If there is such a such as x3 , the function can be re-written as 3!x . If there is a degree much higher than the other degrees in the function, be wary of . If there are any functions of different degrees, one may divide them according to their degree. Step 5: If one has a variable multiplied by a function surrounding another variable, the degrees are added (this is only if the limit is in indeterminate form after all of the other parts have been used). One may make variable exponents become variables with coefficients by taking the of the base and making the variable exponent a coefficient to the logarithm of the base. After one is finished simplifying the function in that way, one must raise both the base and the coefficient to the exponent place of e if both the base and the coefficient contain a variable

3 lim sin (x) x→0 sin(x)−tan(x)

2 lim sin (x) x→0 (x)−(x)

2 lim sin (x) x→0 −x lim 2! x→0 −1 -2 This uses the above stated methodology.

How it works and what it is: Taylor Expansions A Taylor is the sum of an infinite series of differential equations. A Taylor Expansion is a series that approximates how a function behaves around a given number. It does this by summing its equation, one can find the limit. This is partially because of L’Hospital’s rule because it is a series of . This only works for analytical functions. The definition, for ex , is: f(x) = ex

x2 x3 f(x) = 1 + x + 2! + 3!... In a more general form…

f′(a) f′′(a) 2 f′′′(a) 3 f(x) = f(a) + 1! (x − a) + 2! (x − a) + 3! (x − a) ... NOTE: 0! is 1. This series goes on infinitely, in the same pattern. This series can be used to find the limits of certain functions. For example, if one has a function sin(x) approaching a number, we can assume that the will let the sin(x) approach 1 in the case of zero, or the number that the limit is in the case of any other number. This happens because the Taylor expansions are in themselves limits. This is just taking them to an algebraic level and combining them with other methods, so that they will work for all functions. This works for all analytical functions. L’Hopital’s This proposed limit evaluation method can be partially derived from L’Hopital’s rule. L’Hopital’s rule is the idea of differentiating the numerator and denominator of a function to get the limit. In doing that, for a function such as x3 , the function can be re-written as 3!x if there are only functions of powers at or lower than one. In the case that there are functions that are x2 , the function can be re-written as 3x2 . If there is a power much higher than the other powers in the function, be wary of divergence. Basically, this is an algebraic version of L’Hopital’s rule combined with the Taylor series and logarithms. The concept of using L’Hospital’s algebraically and combining it with the Taylor Series and other methods to make a new method can save some time. Logarithms and Simplification In the event that there is a function such as 2x , or a variable in the exponent, one may take the natural logarithm of the function to bring the exponent x down so that the ln (2) is now the coefficient of the x. If one can simplify before using this idea, do so. One may have to raise the solution to an exponent with a base of e if the x was both a base and exponent.

Proof: Assumptions: The limit is not divergent. The coefficients of the numerator are represented as m, and are represented as n in the denominator. This function is real and differentiable on the interval that is being tested, and the of the function g(x) is not zero. It is important to note that, as used in this context, f(x) and g(x) are NOT functions containing multiple x’s such as x2 − x − 2 . They are each individual terms in that equation. f(x)z lim(d→y) g(x)

f (a) f (a) 2 f (a) 3 z!(f(a)+ ′ (x−a)+ ′′ (x−a) + ′′′ (x−a) ...) lim(d→y) 1! 2! 3! g′(a) g′′(a) 2 g′′′(a) 3 g(a)+ 1! (x−a)+ 2! (x−a) + 3! (x−a) ...

f′(a) f′′(a) 2 f′′′(a) 3 a!(lim(f(a)+ 1! (x−a)+ 2! (x−a) + 3! (x−a) ...)) d→y g′(a) g′′(a) 2 g′′′(a) 3 lim(g(a)+ 1! (x−a)+ 2! (x−a) + 3! (x−a) ...) d→y

f (y) f (y) f (y) a!(lim(f(y)+ ′ (d−y)+ ′′ (d−y)2+ ′′′ (d−y)3...)) d→y 1! 2! 3! g (y) g (y) g (y) lim(g(y)+ ′ (d−y)+ ′′ (d−y)2+ ′′′ (d−y)3...) d→y 1! 2! 3!

a!(my) (ny)

lim(x→y)(f(x)g(x)) lim(x→y)g(x)ln(f(x) )

g′(a) g′′(a) 2 g′′′(a) 3 f (a) f (a) f (a) ln( f(a) + ′ (x a) + ′′ (x a)2 + ′′′ (x a)3...) lim(d→y)(g(a) + 1! (x − a) + 2! (x − a) + 3! (x − a) ...) 1! − 2! − 3! −

g′(y) g′′(y) 2 g′′′(y) 3 f(y)) f′′(y) 2 f′′′(y) 3 lim(d→y)(g(y) + 1! (d − y) + 2! (d − y) + 3! (d − y) ...) ln( f(y) + 1! (d − y) + 2! (d − y) + 3! (d − y) ...)

g′(y) g′′(y) 2 g′′′(y) 3 f(y)) f′′(y) 2 f′′′(y) 3 lim(d→y)(g(y) + 1! (d − y) + 2! (d − y) + 3! (d − y) ...) ( f(y) + 1! (d − y) + 2! (d − y) + 3! (d − y) ...)

(g(y)+ g′(y) (d y)+ g′′(y) (d y)2+ g′′′(y) (d y)3...) (f(y)+ f(y)) (d y)+ f′′(y) (d y)2+ f′′′(y) (d y)3...) lim(d→y)(e 1! − 2! − 3! − 1! − 2! − 3! − )

(e(my)(ny) ) (If y is zero, one substitutes 1 in for y) This is the Taylor Series Expansion simplified to an algebraic form.

sinh(tan(ln(ex))) 1 lim x = = 1 x→0 sin(tanh(ln(e ))) 1

(In the case that a is much higher than the power of g(x), then one notes this and investigates zero (if in the numerator) or divergence (if in the denominator) and follows the other instructions.) A Taylor Series is the sum of an infinite series of differential equations. L’Hospital’s rule works by taking the derivative of a function in the numerator and the function in the denominator. Therefore, these two are related. They can be combined to create a new way of evaluating limits. If there is function over a different function, as the fraction approaches any number, L’Hospital’s may be unable to solve it easily if it is nested. If we use the Taylor Series, we get an infinite expansion. The concept of the Taylor Series, however, says that the x will approach the limit in the function, or in the case of zero, one. We can use that in this case to let the x approach that number in the numerator and in the denominator and multiply that by any coefficients that the x has. We can remove any functions surrounding the x, it goes by the Taylor Series that we can do this. If there is a degree, due to L’Hospital’s rule, we can take the exponent’s factorial and make it a coefficient to the x, with additions depending on how high the power is. If there is a fraction, one may take the factorial of both the numerator and denominator. If there is a number with x in the exponent, one may use logarithms to make the base become coefficient to the x. Be careful, as one may have to raise the solution to an exponent with a base of e if the x was both a base and exponent.

Examples: Example 1:

ex−1 1−0 lim(x→0) sin(x) = 1 = 1

This is a simple example. This can be solved by using the proposed methodology. The ex function goes to 1, as the x is substituted back in and the -1 goes to zero. The sin(x) function goes to x (in this case 1), because of Taylor Expansion. Example 2:

−1 x lim(x→0) (sin (tanh(ln(e )))) = 1 = 1 (sinh(tan−1(ln(ex)))) 1 This is a more complex function. It is used to demonstrate how much time can be saved if one is to use this proposed methodology. It also works with a type of Taylor Expansion that all trigonometric functions have in common- the number that the x approaches is the number the function goes to (except for 0. With 0, it goes to 1). Example 3:

2x−1 lim(x→0) x ln(2) 1 ln(2) To make the x have a coefficient, one uses the natural logarithm to turn the base 2 into a coefficient for x. Example 4:

lim(x→0) x √x+4−2

√x+4 1 2 4

1 One must take the square root function, take the 2 power that it is, and put that in the denominator. The x goes to 1, and the 2 goes to zero. Put the √x + 4 in the numerator and solve. Example 5:

1 − 1 lim(x→0) 3+x 3 (x+3)2−9

1 lim(x→0) x3+9x2+27x+27

1 − 54 One must expand the denominator after getting rid of all constants and simplifying. Then, one must take the of the exponents as coefficients to the x’s, add them up, and add three in the place of the eliminated constant (because x=3, and it was not a constant at first). One may then switch the signs, as there was originally subtraction in all of the functions. Example 6:

2 lim(x→2) x −2x+4 x−2 lim(x→y)2x−2 lim(x→y)x

2(2)−2 1 2 For this function, a simple exponent factorial and the x’s moved into ones is all that it takes. Example 7:

sin(x)3 lim(x→��) π−x 0 The power is so many steps above the other powers in the equation that it goes to zero. Example 8:

sin(x) tan(x)

-1 Due to the nature of trigonometric functions, if they are approaching any number except for zero, there will be a sign change from the expected value to the negative/positive of that expected value if there are only trigonometric functions in the problem. With the proposed method, be careful to check for divergence before using it as this method does NOT check for divergence. Otherwise, if one is sure that it does not diverge, one may use the proposed methodology that would work by combining the Taylor Series and L’Hopital’s rule. Example 9:

4 sin(x)√x

√4 x ln(sin(x))

4 e√xsin(x)

e0 1

Conclusion: This proposed methodology removes the need to use calculus in the evaluation of certain indeterminate limits. It can be useful, and can save the user considerable time.