Indeterminate Form
An indeterminate form is of the form 0/0, 0·∞, ∞/∞, etc. These forms occur mostly when we evaluate limits by substituting the given number into the given expression. For example, to evaluate the limit lim _{x → }_{2} (x^{2}  4) / (x  2), we usually substitute x = 2 in the given expression and that gives (2^{2}  4) / (2  2) = 0/0, and the value of 0/0 cannot be determined anymore. Such forms are known as indeterminate forms.
Let us learn more about indeterminate forms and how to solve limits with indeterminate forms.
1.  What are Indeterminate Forms? 
2.  Evaluating Indeterminate Forms of Limits 
3.  Indeterminate Form Table 
4.  FAQs on Indeterminate Form 
What are Indeterminate Forms?
The indeterminate forms are forms whose value cannot be determined. In the process of evaluating some limits, the substitution of the given value of the variable into the given expression cannot provide enough information about the limits. Instead, if that produces the forms such as 0/0, ∞  ∞, ∞/∞, etc, then such forms are known as indeterminate forms. There are 7 indeterminate forms:
 0/0
 0^{0}
 ∞/∞
 0 × ∞
 ∞  ∞
 1^{∞}
 ∞^{0}
These indeterminate forms are a combination of a maximum of two of 0, 1, ∞. The value of each of these forms is not determined because they lead to contradictions when we try to determine their values. Let us just them along with the reasons why they are called indeterminate forms.
Indeterminate Form 0/0
Let us start dividing 0 by 0 and we can find multiple answers. i.e.,
 0 × 1 = 0 ⇒ 0/0 = 1
 0 × 2 = 0 ⇒ 0/0 = 2
 0 × 3 = 0 ⇒ 0/0 = 3
and so on
It means, 0/0 = 1 = 2 = 3 = .... It means 1 = 2 = 3 = ... which is absolutely a contradiction. It means the value of 0/0 cannot be determined and hence it is an indeterminate form. We can understand this from the figure below.
Indeterminate Form 0^0
We can calculate 0^{0} in two ways.
 We know that a^{0} = 1 for any 'a'. Using this 0^{0} = 1.
 We know that 0^{2} = 0 × 0 = 0 ; 0^{3} = 0 × 0 × 0 = 0, etc. From this, 0^{0} = 0
From the above arguments 0^{0} = 1 = 0. This lead to 1 = 0, which is a contradiction. Hence 0^{0} is an indeterminate value.
Indeterminate Form ∞/∞
We already proved that 0/0 is an indeterminate form. Also, we know that 1/∞ = 0. We can write 0/0 as
0/0 = (1/∞) / (1/∞)
= 1/∞ × ∞/1
= ∞/∞
It means ∞/∞ is an equivalent form of 0/0 and hence is an indeterminate form.
Indeterminate Form 0 × ∞
We will again use the same 0/0 form to prove this. We already know that 1/0 = ∞. Now,
0/0 = 0 × 1/0 = 0 × ∞
Since 0/0 is an indeterminate form, 0 × ∞ is also an indeterminate form.
Indeterminate Form ∞  ∞
∞ represents a very very large number. But ∞  ∞ doesn't mean that it is "the number minus the same number" because ∞ doesn't represent any fixed number. So we cannot say that ∞  ∞ = 0, rather, its value cannot be determined. Thus, ∞  ∞ is an indeterminate form.
Indeterminate Form 1^∞
Of course, multiplying 1 by itself for an infinite number of times leads to 1. But still, 1 raised to ∞ is an indeterminate form. For this, we need a little more analysis.
 If we take a number that is less than and very close to 1, then multiplying it by itself an infinite number of times, gives a very very small number and is approximately equal to 0.
 If we take a number that is greater than and very close to 1, then multiplying it by itself an infinite number of times, gives a very very bigger number and is approximately equal to ∞.
It means, the limit lim _{x → }₁ x^{∞} does not exist because its lefthand limit is 0 and the righthand limit is ∞. Hence, if we get 1^{∞} after the substitution into the limit, it means that we have got an indeterminate form.
Indeterminate Form ∞^0
We know that a^{0} using the quotient rule of exponents can be written as a/a. In the same way, ∞^{0} can be written as ∞/∞, which is an indeterminate form. Thus, ∞^{0} is an indeterminate form.
Evaluating Indeterminate Forms of Limits
Though there are 7 indeterminate forms, the most occurring indeterminate forms are 0/0 and ∞/∞. For calculating limits leading to these forms, L'Hopital's rule is most helpful. But sometimes, other methods are also applicable. Let us discuss each method along with examples.
Factoring Method
Whenever we get an indeterminate form while evaluating limits by direct substitution, the easiest way is to see whether factoring allows any cancellation of terms. Sometimes, the direct substitution after the cancellation of common factors (from numerator and denominator) would prevent getting indeterminate forms. Here is an example.
Example: As we discussed in the beginning if we substitute x = 2 in the limit lim _{x → }_{2} (x^{2}  4) / (x  2), we get 0/0. Now we will factor the numerator using a^{2}  b^{2} formula. Then we get x^{2}  4 = x^{2}  2^{2} = (x + 2)(x  2). Then:
lim _{x → }_{2} (x^{2}  4) / (x  2) = lim _{x → }_{2} [(x + 2)(x  2)] / (x  2)
= lim _{x → }_{2} (x + 2)
= 2 + 2
= 4
Hence, we could evaluate the limit by factoring and canceling the common terms.
Taking the Highest Power Term as Common Factor
Most of the times, the limits like lim _{x → ∞} (2x^{2}  4x + 1) / (3x^{2}  8x + 3) would tend to ∞/∞ by direct substitution x = ∞. In such cases, we can take the highest power term (which is x^{2} in each of the numerator and denominator in this case) as the common factor and simplify it.
lim _{x → ∞} (2x^{2}  4x + 1) / (3x^{2}  8x + 3)
= lim _{x → ∞} [x^{2 }(2  4/x + 1/x^{2})] / [x^{2 }(3  8/x + 3/x^{2})]
= lim _{x → ∞} (2  4/x + 1/x^{2}) / (3  8/x + 3/x^{2}) (x^{2} has got cancelled)
= (2  0 + 0) / (3  0 + 0)
= 2/3
L'Hopital's Rule
L'Hopital's rule is very helpful in evaluating the limits with indeterminate forms. This rule says to take the derivative of numerator and denominator separately and then apply the limit whenever we get an indeterminate form by direct substitution. If we get an indeterminate form after the first application of L'Hopital's rule, then the rule can be applied again. Here is an example.
Example: lim _{x → }_{0} (sin x / x) = 0/0 by direct substitution of x = 0. Now, by L'Hopital's rule, this limit is same as
lim _{x → }_{0} (cos x / 1) (as derivative of sin x is cos x; and the derivative of x is 1)
= (cos 0) / 1
= 1/1
= 1
This rule can be applied instead of the above two methods (of "factoring method" and "taking the highest power term as common factor") as well.
Taking ln in Cases of 1^{∞}, 0^{0}, and ∞^{0}
Whenever we get an indeterminate form with exponents such as 1^{∞} while evaluating a limit, then assume that limit as L and apply the natural logarithm "ln" on both sides. But don't forget to convert the log form into exponential form at the end to calculate L. Here is an example:
Example: Evaluate the limit lim _{x → }_{0} (1 + x)^{1/x}.
Solution:
Let L = lim _{x → }_{0} (1 + x)^{1/x}
Taking "ln" on both sides,
ln L = lim _{x → }_{0} (1/x) ln (1 + x)
= lim _{x → }_{0} [ ln (1 + x) ] / x
= lim _{x → }_{0} [1/(1+x)] / 1 (by L'Hopital's rule)
= 1/(1+0)
= 1
So we got ln L = 1. From this, L = e^{1} = e.
Indeterminate Form Table
Sometimes, a combination of the above methods also would work in evaluating the limits with indeterminate forms. The following table gives each of the 7 indeterminate forms along with an example. Also, each example is solved for you in an alternative method.
Indeterminate Form  Example (Limit with Direct Substitution) 
Evaluating Limit 

0/0  lim _{x → }₁ (x^{2}  1) / (x  1) = 0/0  Factor method (or) L'Hopital's rule lim _{x → }₁ (x^{2}  1) / (x  1) = lim _{x → }₁ [(x  1)(x + 1)] / (x  1) = lim _{x → }₁ (x + 1) = 1 + 1 = 2 
1^{∞}, 0^{0}, and ∞^{0}  lim _{x → }_{0} x^{x} = 0^{0}  Assume the limit as L and apply "ln": Let L = lim _{x → }_{0} x^{x} ln L = lim _{x → }_{0} x ln x ln L = lim _{x → }_{0} (ln x) / (1/x) Using L'Hopital's rule, ln L = lim _{x → }_{0} (1/x) / (1/x^{2}) ln L = lim _{x → }_{0} (x) ln L = 0 = 0 L = e^{0} = 1 
∞/∞  lim _{x → ∞} (3x + 1) / (2x + 3) = ∞/∞  Taking highest power term as common factor (or) L'Hopital's rule lim _{x → ∞} [x (3 + 1/x) ] / [x (2 + 3/x)] = lim _{x → ∞}(3 + 1/x) / (2 + 3/x) = (3 + 0) / (2 + 0) = 3/2 
∞  ∞  lim _{x →} _{π/2}^{ }(tan x  sec x) = ∞  ∞  Write as fraction and apply L'Hopital's rule lim _{x →} _{π/2} [ (sin x/cos x)  (1/cos x)] = lim _{x →} _{π/2} [ (sin x  1) / cos x] By L'hopital's rule = lim _{x →} _{π/2} (cos x) / ( sin x) = lim _{x →} _{π/2} ( cot x) =  cot π/2 = 0 = 0 
Important Notes on Indeterminate Form:
 1/0 is not an indeterminate form, rather 1/0 = ∞.
 0/1 is not an indeterminate form, its value is 0/1 = 0.
 1/∞ is not an indeterminate form, it is 1/∞ = 0.
 ∞/1 is not an indeterminate form, but ∞/1 = ∞.
☛ Related Topics:
Indeterminate Form Examples

Example 1: Evaluate the limit lim _{x → }_{0} (sin 6x/ x).
Solution:
If we substitute x = 0 in the given expression, we get (sin 0)/0 = 0/0, which is an indeterminate form. So we apply L'Hopital's rule.
lim _{x → }_{0} (sin 6x / x) = lim _{x → }_{0} (6 cos 6x) / 1
= 6 cos (6 × 0)
= 6 cos 0
= 6 (1)
= 6Answer: lim _{x → }_{0} (sin 6x / x) = 6.

Example 2: Find the value of lim _{x → ∞} x^{1/x}.
Solution:
If we apply the limit as x → ∞, then we get the exponent indeterminate form ∞^{0}.
So we assume the given limit to be L and apply "ln".
Let L = lim _{x → ∞} x^{1/x}
Applying "ln" on both sides,
ln L = lim _{x → ∞} [(1/x) ln x]
= lim _{x → ∞} ( ln x / x)This would give us the indeterminate form ∞/∞. So we apply L'Hopital's rule.
= lim _{x → ∞} [ (1/x) / 1]
= 1/∞
= 0Thus, ln L = 0 which gives L = e^{0} = 1.
Answer: lim _{x → ∞} x^{1/x} = 1.

Example 3: What is the value of the limit lim _{x → ∞} (x^{3}/e^{x})?
Solution:
The direct application of the limit to the given expression gives the indeterminate form ∞/∞. Apply L'Hopital's rule,
lim _{x → ∞} (x^{3}/e^{x}) = lim _{x → ∞} (3x^{2}/e^{x})
This would again give ∞/∞. Apply L'Hopital's rule again,
= lim _{x → ∞} (6x/e^{x})
We apply L'Hopital's rule again as it is again giving ∞/∞.
= lim _{x → ∞} (6/e^{x})
= 6/e^{∞}
= 6/∞
= 0Answer: lim _{x → ∞} (x^{3}/e^{x}) = 0.
FAQs on Indeterminate Form
What is the Meaning of Indeterminate Form?
An indeterminate form is a mathematical expression whose value cannot be determined. When we try to determine its value, it leads to a contradiction. Examples: 0/0, 0^{0}, ∞/∞, etc are indeterminate forms. If we try to find what is 0/0, then we can get infinite values like 1, 2, 3, .... because 0 multiplied by any number is 0 itself and this gives 0^{0} = 1 = 2 = 3 = ... which is a contradiction as all numbers cannot be equal to each other.
What are 7 Indeterminate Forms of Limits?
There are 7 indeterminate forms which are mentioned as follows:
 0/0
 ∞/∞
 0^{0}
 1^{∞}
 ∞^{0}
 0 × ∞
 ∞  ∞
How do I Remove Indeterminate Form?
We come across indeterminate forms while evaluating limits by substitution of the given value. In such cases, we apply one of the following methods:
 factoring (and canceling)
 take the highest power term as the common factor
 apply L'hopital's rule
 assume the given limit as L and apply "ln" on both sides.
Sometimes, we would need to apply a combination of these methods as well.
How to Calculate Indeterminate Forms of Limits?
 To calculate limits with indeterminate forms such as 0/0, ∞/∞, ∞  ∞, and 0 × ∞, writing the given function as a fraction and applying L'Hopital's rule would be the better option.
 To evaluate the limits that lead to indeterminate forms of exponents such as 0^{0}, 1^{∞}, and ∞^{0} we assume the given limit as a variable (say L) and then apply "ln" on both sides.
Is Infinity^Infinity an Indeterminate Form?
No, ∞^{∞} is NOT an indeterminate form, it is a determinate form but its value is ∞ (undefined). On the other hand, ∞  ∞ is an indeterminate form.
Is Indeterminate Form Same as Undefined?
No, indeterminate form and undefined values are different.
 The indeterminate values are 0/0, ∞/∞, 0^{0}, 1^{∞}, ∞^{0}, 0 × ∞, and ∞  ∞. These values when tried to be found lead to contradictions.
 Undefined values mean that are not defined. For example, 1/0, 1/0, ∞, and ∞ are undefined values. These values cannot be found.
What Do You Do with Indeterminate Forms of Limits?
Indeterminate forms mostly occur while evaluating limits. In such cases, we go for factoring (or) L'hopital's rule. In the case of indeterminate forms of exponents, we apply the natural logarithm "ln".
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