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Indeterminate Forms

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Last updated date: 25th Apr 2024
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Introduction of Indeterminate Forms

The indeterminate form usually involves two fractions whose limit cannot be determined by referring to the original limits of the two individual functions. Such types of functions are common in calculus. The limit of the derivative in this regard tends to become the limit of an indeterminate form. In this article, we shall discuss the indeterminate forms list, indeterminate forms examples, along with discussing how to evaluate indeterminate forms. Now that you understand what is an indeterminate form, let us move on to the limiting behavior of such functions. 


There are a few types of limits that are also regarded as indeterminate since, due to an individual part of the expression, it is difficult to calculate the overall limit of the expression.


For instance, consider that the expression given is as follows:


\[\lim_{x\to\infty}f(x)\;=\;\lim_{x\to\infty}g(x)\;=\;0\;then\lim_{x\to\infty}\frac{f(x)}{g(x)}\]


In this case, if the limits are used on the given function, then the resultant becomes ∞/∞, which could be referred to as an indeterminate form. All indeterminate forms are bound to include any of seven of the following expressions, which make the expression an indeterminate form. These are as follows:


∞/∞, ∞ -∞, 0/0,00, 1 , ∞0 , 0 x ∞


What is the Importance of Indeterminate Forms in Calculus?

Indeterminate forms are embedded all over calculus. When we are dealing with ratios like 1/0 and 0/0, or with an infinity of any form, we will most likely require the use of another theorem, such as L’Hopital’s Rule, in order to solve for a limit. Since it is known that indeterminate forms are often solvable can shed some light upon a function that was previously unavailable.


If we dig further into the nuances of indeterminate forms, it allows the utilization of a variety of ways for determining the characteristics of any function at any particular limit. An indeterminate form is determined by the lack of sufficient information. However, sometimes, there is too much information that needs to bring down a solution to a single technique.


Consider the fraction 0/0. Anybody with elementary knowledge would think that this fraction is equal to one since the numerator matches the denominator, which is supposed to give one. On a more practical approach, one could assume the fraction to be simply zero. Moreover, a zero in the denominator may represent infinity or non-existent. Hence, there are different ways to classify this ratio, but there is no clear winner in this.


Then why would indeterminate forms be necessary in pragmatic situations in real life? The obvious answer to this question is that these are not essential, at least not by themselves. But their presence builds a foundation for us to come up with different ways for making indeterminate forms solvable.


Various areas in maths and physics prefer the use of a denominator of zero as a potential starting point in various situations and formulas. Laws that involve physics, heat, and quantum mechanics generally use indeterminate forms at some point or another. An understanding of these forms will enable you to solve them using a proper method, be it L’Hopital’s Rule or any other method.


If we take a more practical standpoint, some very common problems involving time, motion, and velocity, requiring the use of indeterminate forms, could be there. Problems relating to the speed and time of moving objects will need to begin at zero. An indeterminate form will be needed to determine an infinite limit of two of such objects.

It is important to understand the cause behind classifying an indeterminate form as “indeterminate”. This is done to separate it from different ratios, which can be zero or that do not exist. It is a possibility that the substitution of the word “temporary” for indeterminate may clear up some of the misconceptions around the use of such forms.


Considering these forms to be ephemeral can prove to be a better way of thinking of them. An indeterminate form can thus be taken as just a tool for further calculation, which is left at the discretion of the consumer of this information.


List of all Indeterminate Forms

Now that we know that there are seven types of indeterminate form, let us look at the conditions that make them so:


Case 1: 0/0

Conditions: \[\lim_{x\to c}f(x)=0,\;\lim_{x\to c}g(x)=0\]


Case 2: ∞/∞

Conditions: \[\lim_{x\to c}f(x)=\infty,\;\lim_{x\to c}=g(x)=\infty\]


Case 3:  0 x ∞

Conditions:\[\lim_{x\to c}f(x)=0,\;\lim_{x\to c}=g(x)=\infty\]


Case 4: ∞ -∞

Conditions: \[\lim_{x\to c}f(x)=1,\;\lim_{x\to c}=g(x)=\infty\]


Case 5: 00

Conditions: \[\lim_{x\to c}f(x)=0,\;\lim_{x\to c}=g(x)=0\]


Case 6: 1

Conditions: \[\lim_{x\to c}f(x)=\infty,\;\lim_{x\to c}=g(x)=0\]


Case 7:0

Conditions: \[\lim_{x\to c}f(x)=\infty,\;\lim_{x\to c}=g(x)=\infty\]


Methods to Evaluate Indeterminate Forms 

  1. Factoring Method - This method is usually used for 0/0 form and involves factoring the expressions given to their simplest form. After having derived the simplest form, the limit value is used to solve.

  2. L'Hospital's Rule - The rule states that in the event of an indeterminate form, the way to solve it would be to differentiate the numerator and the denominator separately and then apply the limit. In this method, the derivatives of the numerator and denominator are considered individually after each step to see if they become free of the variable thereby making at least one of the terms constant.

  3. Division by Highest Power - This method is usually employed when the indeterminate form is usually given in the ∞/∞ format. The way to proceed in such a case would be to divide both the numerator and the denominator of the given expression by the variable of the highest power in the sum. Subsequently, the limit value is obtained after this.


Solved Examples 

  1. \[\lim_{x\to 0}\frac{sinx-x}{x^{3}}=\lim_{x\to 0}\frac{Cosx-1}{3x^{2}}\]


                         = \[\lim_{x\to 0}\frac{-sinx}{6x}\]


                         = \[\lim_{x\to 0}\frac{-cosx}{6}=\frac{-1}{6}\]

 

  1. \[\lim_{x\to 0}\frac{2cosx-2+x^{2}}{x^{4}}=\lim_{x\to 0}\frac{-2sinx+2x}{4x^{3}}\]


  = \[\lim_{x\to 0}\frac{-2cosx+2}{12x^{2}}\]


    = \[\lim_{x\to 0}\frac{2sinx}{24x}=\lim_{x\to 0}\frac{2cosx}{24}=\frac{2}{24}=\frac{1}{12}\]

 

  1. \[\lim_{x\to \infty}\frac{ln(3^{x}+2^{x})}{x}=\lim_{x\to \infty}\frac{3^{x}ln2+2^{x}ln2}{3^{x}+2^{x}}\]


             = \[\lim_{x\to \infty}\frac{ln3+(2/3)^{x}ln2}{1+(2/3)^{x}}\]


= \[\frac{ln3+0}{1+0}=ln3\]

 

  1. \[\lim_{x\to \infty}\frac{x^{2}}{e^{x}}=\lim_{x\to \infty}\frac{2x}{e^{x}}\]


          = \[\lim_{x\to \infty}\frac{2}{e^{x}}\]


          = \[\frac{2}{\infty}=0\]

 

  1.  \[\lim_{y\to 2}\frac{y^{3}+3y^{2}+2y}{y^{2}-y-6}=\lim_{y\to 2}\frac{y(y+1)(y+2)}{(y-3)(y+2)}\]


                    = \[\lim_{y\to 2}\frac{y(y+1)}{y-3}\]


                    = \[\frac{\lim_{y\to -2}y\;.\;\lim_{y\to -2}(y+1)}{\lim_{y\to -2}(y-3)}\]


                    = \[\frac{-2\;.\;(-1)}{-5}=-\frac{2}{5}\]

FAQs on Indeterminate Forms

1. Can L’Hospital’s Rule be applied in the case of an expression where the numerator or the denominator is a finite non-zero limit?

No,  L'Hospital's rule cannot be applied in the case of an expression where the numerator or the denominator is a finite non-zero limit. To understand why this happens, it is important to first understand the indeterminate form definition. Once you are familiar with the concept, you shall see that there is a particular type of indeterminate -  ∞/∞, in which case the L’Hospital’s rule is applied. The general inference here is that usually, using the rule in the case of an expression with finite non-zero limit tends to give wrong answers. 

2. What is an equivalent infinitesimal?


Consider two variables, α, and β. When these two variables converge at the same point and lim β/α = 1, these variables are usually referred to as equivalent infinitesimal. This is usually represented mathematically as α ∼ β. 


Subsequently, if the above-used analogy is extended to mathematical conditions such that α ∼ α’ as well as β ∼ β’, and if these conditions are prevalent, then the following is true:


\[lim \frac{\beta}{\alpha} = lim \frac{\beta'}{\alpha'}\]