Indeterminate Forms

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 Example along with 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, if the expression given is:

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

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 ∞

 

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

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

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

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


FAQ (Frequently Asked Questions)

1. Can L Hospital’s Rule Be Applied in Case of An Expression Where the Numerator or the Denominator is a Finite Non Zero Limit?

Ans: No,  L'Hospital's rule cannot be applied in 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 rule is applied. The general inference here being that usually using the rule in case of an expression with finite non zero limit tends to give wrong answers. Here’s an example of the same: 

Consider the sum:

Now if we were to not use L'Hospital's rule blindly, then the result would be: 

However, a use of L'Hospital's rule will give us the answer as: 

2. What is An Equivalent Infinitesimal?

Ans: 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’, if these conditions are prevalent then:

lim β/α = lim β’/α’

The mathematical proof for such an occurrence is not really important however, it is given below just for the sake of reference.