Indeterminate Forms of Limits

What Are Indeterminate Forms?

When we try to calculate a limit and direct substitution doesn’t give a clear result, we encounter an indeterminate form. In simple words, these are expressions where the value of the limit cannot be directly determined. For example, the limits $\lim_{x \to 0} \frac{\sin x}{x}$ or $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ initially give expressions like $\frac{0}{0}$. Such cases are called indeterminate forms of limits because they can take different values depending on how the function behaves near that point.

Common signs of indeterminate forms include:

• $\frac{0}{0}$: Both numerator and denominator approach 0

• $\frac{\infty}{\infty}$: Both numerator and denominator grow very large

• $0 \cdot \infty$: One factor goes to 0, the other to infinity

• $\infty - \infty$: Difference of two very large numbers

• $0^0$, $1^\infty$, $\infty^0$: Indeterminate exponential forms

Recognizing these common indeterminate forms in calculus is the first step to solving limit problems effectively. Once identified, we can apply techniques like L’Hospital’s Rule, algebraic simplification, or rationalization to find the exact limit.

Limit of Indeterminate Form and Algebraic Limit

Indeterminate forms arise when evaluating limits and direct substitution does not give a specific value. If we try to evaluate $\lim\limits _{x \rightarrow a} f(x)$ and get one of the following forms:

$\frac{0}{0}, \frac{\infty}{\infty}, \infty - \infty, 1^\infty, 0^0, \infty^0, 0 \cdot \infty$

then the limit is called an indeterminate form.

Examples:

1. $\lim\limits _{x \rightarrow 2} \frac{x^2-4}{x-2} = \frac{0}{0}$

2. $\lim\limits _{x \rightarrow 0} \frac{\sin x}{x} = \frac{0}{0}$

3. $\lim\limits _{x \rightarrow \pi / 2} (\tan x)^{\cos x} = \infty^0$

Note:

• The $\frac{0}{0}$ form means both numerator and denominator approach 0, but not necessarily exactly 0.
  Example: $\lim\limits _{x \rightarrow 0^+} \frac{[x]}{x}$ → here the denominator tends to 0, but numerator is exactly 0, so it is not a $\frac{0}{0}$ form.

• One indeterminate form can sometimes be converted into another.

L’Hospital’s Rule

Definition: If $\lim\limits _{x \rightarrow a} \frac{f(x)}{g(x)}$ is of $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, then:

$\lim\limits _{x \rightarrow a} \frac{f(x)}{g(x)} = \lim\limits _{x \rightarrow a} \frac{f'(x)}{g'(x)}$

If the indeterminate form still persists after differentiation, differentiate again:

$\lim\limits _{x \rightarrow a} \frac{f(x)}{g(x)} = \lim\limits _{x \rightarrow a} \frac{f'(x)}{g'(x)} = \lim\limits _{x \rightarrow a} \frac{f''(x)}{g''(x)}$

Continue this process until the form is no longer indeterminate. This is the L'Hospital's form.

Applications of L’Hospital’s Rule:

1. $\lim\limits _{x \rightarrow 0} \frac{\sin x}{x} = \lim\limits _{x \rightarrow 0} \frac{\cos x}{1} = 1$

2. $\lim\limits _{x \rightarrow \infty} \frac{\ln x}{x} = \lim\limits _{x \rightarrow \infty} \frac{1/x}{1} = 0$

3. $\lim\limits _{x \rightarrow 0} \frac{\ln(1+x)}{x} = \lim\limits _{x \rightarrow 0} \frac{1/(1+x)}{1} = 1$

Indeterminate Forms of Limits: Complete List with Conditions, Transformations, and Examples

This section provides a complete list of indeterminate forms of limits, explaining their conditions, step-by-step transformations, and solved examples to help students quickly understand and evaluate tricky limit problems.

1. $\frac{0}{0}$ Form

Condition: Occurs when both the numerator and the denominator of a fraction approach zero as the independent variable approaches a certain value.
Transformation: To evaluate a $\frac{0}{0}$ limit, use techniques like:

• Factorization: Factor the numerator and/or denominator to cancel common factors.

• Rationalization: Multiply by a suitable conjugate or form of 1 to simplify the expression.

• L’Hospital’s Rule: Differentiate the numerator and denominator separately, if the limit exists.
  Example:
  $\lim\limits _{x \rightarrow 0} \frac{\sin x}{x} = \frac{0}{0}$
  Applying L’Hospital’s Rule:
  $\lim\limits _{x \rightarrow 0} \frac{\sin x}{x} = \lim\limits _{x \rightarrow 0} \frac{\cos x}{1} = 1$

2. $\frac{\infty}{\infty}$ Form

Condition: Occurs when both the numerator and the denominator approach infinity as the variable approaches a certain value.
Transformation: To resolve this form:

• L’Hospital’s Rule: Useful when both numerator and denominator are differentiable near the point of interest.

• Rewriting/Fraction Simplification: Factor or manipulate terms to convert the fraction into a simpler, solvable form.

3. $0 \cdot \infty$ Form

Condition: Appears when a limit involves the product of a term approaching zero and another approaching infinity.
Transformation: Convert it into a $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form using:

• Algebraic Manipulation: Rewrite the product as a fraction or separate terms.

• Example:
  $\lim\limits _{x \rightarrow 0} x \cdot \frac{1}{x} = 0 \cdot \infty$
  Rewriting: $\lim\limits _{x \rightarrow 0} 1 = 1$

4. $\infty - \infty$ Form

Condition: Occurs when a limit involves the difference of two terms, each approaching infinity.
Transformation: Transform the expression to a more manageable form:

• Algebraic Manipulation: Combine terms or simplify to get $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

• Factorization: Rearrange and factor terms to simplify evaluation.
  Example:
  $\lim\limits _{x \rightarrow \infty} (x + \sqrt{x}) - x = \infty - \infty$
  Simplifying: $\lim\limits _{x \rightarrow \infty} \sqrt{x} = \infty$

5. $0^0$ Form

Condition: Appears when a limit involves an expression like $x^x$ as $x \to 0$.
Transformation: Use logarithmic and exponential techniques:

• Rewrite $x^x = e^{x \ln x}$ and evaluate the limit using known limits of $x \ln x$.

6. $\infty^0$ Form

Condition: Occurs when a limit involves an expression like $\infty^0$.
Transformation: Convert the expression using logarithms:

• $y = f(x)^{g(x)} \implies \ln y = g(x) \ln f(x)$, then evaluate the limit using L’Hospital’s Rule if needed.

7. $1^\infty$ Form

Condition: Appears when a limit involves an expression like $1^{\infty}$.
Transformation: Use exponential and logarithmic techniques:

• Rewrite as $y = f(x)^{g(x)} \implies \ln y = g(x) \ln f(x)$, then solve using L’Hospital’s Rule or standard limit methods.

Methods to Evaluate Indeterminate Forms

This section explains the key methods to evaluate indeterminate forms, including algebraic simplification, factoring, rationalization, and L’Hospital’s Rule, with clear examples for easy understanding and quick problem-solving.

Algebraic Simplification

• Simplify fractions, expand expressions, or divide by the highest power.

• Example:
  $\lim\limits_{x \to \infty} \frac{3x^3 + 2x}{5x^3 - x^2} = \frac{\infty}{\infty}$
  Divide numerator & denominator by $x^3$:
  $\lim\limits_{x \to \infty} \frac{3 + \frac{2}{x^2}}{5 - \frac{1}{x}} = \frac{3}{5}$

Factoring and Canceling

• Factor common terms to cancel and remove $\frac{0}{0}$.

• Example:
  $\lim\limits_{x \to 3} \frac{x^2 - 9}{x - 3} = \frac{0}{0}$
  Factor numerator: $\frac{(x-3)(x+3)}{x-3} = \lim\limits_{x \to 3} x+3 = 6$

Rationalization Techniques

• Multiply by conjugate to simplify limits involving roots.

• Example:
  $\lim\limits_{x \to 1} \frac{\sqrt{x+3} - 2}{x-1} = \frac{0}{0}$
  Multiply numerator & denominator by $\sqrt{x+3}+2$:
  $\lim\limits_{x \to 1} \frac{x+3 - 4}{(x-1)(\sqrt{x+3}+2)} = \lim\limits_{x \to 1} \frac{1}{\sqrt{4}+2} = \frac{1}{4}$

Using L’Hospital’s Rule

• Apply when $\frac{0}{0}$ or $\frac{\infty}{\infty}$ forms appear. Differentiate numerator & denominator.

• Example:
  $\lim\limits_{x \to 0} \frac{\sin x}{x} = \frac{0}{0}$
  Using L’Hospital: $\lim\limits_{x \to 0} \frac{\cos x}{1} = 1$

• Example 2:
  $\lim\limits_{x \to \infty} \frac{e^x}{x^2} = \frac{\infty}{\infty}$
  Differentiate: $\lim\limits_{x \to \infty} \frac{e^x}{2x} = \infty$

Indefinite Forms of limits

Solved Examples Based On Limits on Indeterminate Forms:

Example 1: Let $f: R \rightarrow R {\text { be a continuously differentiable function such that }} f(2)=6$ and $f^{\prime}(2)=\frac{1}{48} \int_6^{f(x)} 4 t^3 d t=(x-2) g(x) \lim\limits _6 g(x)$, then $x \rightarrow 2$ is equal to : [JEE Main 2018]

1) $18$

2) $24$

3) $12$

4) $36$

Solution:

L - Hospital Rule -

In the form of $\frac{0}{0}$ and $\frac{\infty}{\infty}$ wedifferentiate $\frac{N^r}{D^r}$ separately.

$
\Rightarrow \lim\limits _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim\limits _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}
$

- wherein

$
\lim\limits _{x \rightarrow a} \frac{\frac{d}{d x} f(x)}{\frac{d}{d x} g(x)}
$

Where $f(x)$ and $g(x)=0$

Evaluation of limits : (algebraic limits) : (Method of direct substitution) -
$\lim\limits _{x \rightarrow a} f(x)$ defines by direct substituting $x=a$

$
e x: \lim\limits _{x \rightarrow 1} \frac{x^2+x+1}{x^2+x-1}=3
$

- wherein

Means at $x=a, f(x)$ defined.

$
\begin{aligned}
f: R \rightarrow R \quad f(2)=6 \\
\begin{aligned}
\lim\limits _{x \rightarrow 2} g(x) & =\lim\limits _{x \rightarrow 2} \frac{\int_6^{f(x)} 4 t^3 d t}{x-2} \\
& =\lim\limits _{x \rightarrow 2} \frac{4 \cdot f^3(x) \cdot f^{\prime}(x)}{1} \\
& =4 f^3(2) f^{\prime}(2)=\frac{1}{48} \\
& =4 \times(6)^3 \times \frac{1}{48} \\
& =6 \times 36 \times \frac{1}{12}=18
\end{aligned}
\end{aligned}
$

Hence, the answer is the option (1).

Example 2: $\lim\limits _{x \rightarrow 1} \frac{\int_0^{(x-1)^2} t \cdot \cos \left(t^2\right) d t}{(x-1) \sin (x-1)}$ is equal to? [JEE Main 2020]
1) $-\frac{1}{2}$
2) $\frac{1}{2}$
3) Does not exist
4) $0$

Solution:
Given,

$
\lim\limits _{x \rightarrow 1} \frac{\int_0^{(x-1)^2} t \cdot \cos \left(t^2\right) d t}{(x-1) \sin (x-1)}
$

Apply L'Hôpital's rule

$
\begin{aligned}
& \Rightarrow \lim\limits _{x \rightarrow 1} \frac{\int_0^{(x-1)^2} t \cdot \cos \left(t^2\right) d t}{(x-1) \sin (x-1)}=\lim\limits _{x \rightarrow 1} \frac{2(x-1)^3 \cos \left((x-1)^4\right)}{\sin (x-1)+(x-1) \cos (x-1)} \\
& \Rightarrow \lim\limits _{x \rightarrow 1} \frac{\int_0^{(x-1)^2} t \cdot \cos \left(t^2\right) d t}{(x-1) \sin (x-1)}=\lim\limits _{x \rightarrow 1} \frac{2(x-1)^2 \cos (x-1)^4}{\frac{\sin (x-1)}{(x-1)}+\cos (x-1)}=0
\end{aligned}
$
Hence, the answer is the option (4).

Example 3:

$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$ equals
1) $
4 \sqrt{2}
$

2) $
\sqrt{2}
$

3) $
2 \sqrt{2}
$

4) $
4
$

Solution:

$
\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}
$

normalizing :

$
\begin{aligned}
& : \frac{\left(\sin ^2 x\right)(\sqrt{2}+\sqrt{1+\cos x})}{(2-1-\cos x)} \\
& =\left(\sin ^2 x\right)=1-\cos ^2 x=(1-\cos x)(1+\cos x) \\
& \lim _{x \rightarrow 0} \frac{(1-\cos x)(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})}{(1-\cos x)} \\
& =\lim _{x \rightarrow 0}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x}) \\
& =(1+1)(\sqrt{2}+\sqrt{1+1}) \\
& =2(2 \sqrt{2}) \\
& =4 \sqrt{2}
\end{aligned}
$
Hence, the answer is the option 1.

Example 4: Which of the following is not indeterminate form when $x$ tends to zero?
$\frac{x+1}{x+2}$
2) $\frac{\sin x}{x}$
3) $\frac{\tan x}{x}$
4) $\frac{1-\cos x}{x^2}$

Solution: On direct substitution (A) gives $1 / 2$ while (B),(C).(D) are of form $\frac{0}{0}$
So $(A)$ is correct
Hence, the answer is the option 1.
Question 5 : Which of the following is not an indeterminate form when $x$ approaches zero?
1) $(1+x)^{1 / x}$
2) $(\cos x)^{1 / x^2}$
3) $(\tan (\pi / 4+x))^{1 / x}$
4) $(1+x)^{x+2}$

Solution:
(1). (2) and (3) all are of the form $1^{\infty}$, while (D) is directly known approaching i.e 1

So (4) is correct
Hence, the answer is the option 4.

List of topics related to Limits

This section provides a list of important limit topics, helping students focus on key concepts and prepare efficiently for exams and competitive tests.

Sandwich Theorem

Limits of Trigonometric Functions

Limits using expansion

Left and Right Hand Limits

Higher Order Derivatives

NCERT Resources

Explore NCERT notes, exemplar solutions, and step-by-step solved examples to master limits and indeterminate forms in a structured and exam-oriented way.

NCERT Notes for Class 11 Maths Chapter 13 - Limits and Derivatives

NCERT Solutions for Class 11 Maths Chapter 13 - Limits and Derivatives

NCERT Exempar Solutions for Class 11 Maths Chapter 13 - Limits and Derivatives

Practice Questions based on Indeterminate forms

This section offers practice problems and solved examples on indeterminate forms, enabling students to apply concepts, improve problem-solving speed, and gain confidence for exams.

Limit Of Indeterminate Form And Algebraic Limit - Practice Question MCQ

We have shared the links below to practice questions on the related topics of limits: