IIT JEE Main Maths -Unit 7- Limits of a Function- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 7- Limits of a Function – Study Notes – New syllabus
IIT JEE Main Maths -Unit 7- Limits of a Function – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Concept of Limit
- Limits of Polynomials and Rational Functions
- Limits Involving Trigonometric and Exponential Functions
- Limits Involving Indeterminate Forms (0/0, ∞/∞, 1^∞, 0⁰, ∞⁰)
Concept of Limit
The concept of a limit is fundamental to calculus. It describes the value that a function approaches as the input approaches a certain number.
$ \lim_{x \to a} f(x) $
means the value that \( f(x) \) approaches when \( x \) gets arbitrarily close to \( a \) (but not necessarily equal to \( a \)).
Formal Definition
\( \displaystyle \lim_{x \to a} f(x) = L \) if for every number \( \varepsilon > 0 \), there exists a number \( \delta > 0 \) such that whenever \( 0 < |x – a| < \delta \), it follows that \( |f(x) – L| < \varepsilon \).
Intuitive meaning: As \( x \) gets closer to \( a \), \( f(x) \) gets closer to \( L \).
Left-Hand and Right-Hand Limits
When analyzing behavior near \( x = a \):
- Left-hand limit (LHL): \( \displaystyle \lim_{x \to a^-} f(x) \) → as \( x \) approaches \( a \) from the left.
- Right-hand limit (RHL): \( \displaystyle \lim_{x \to a^+} f(x) \) → as \( x \) approaches \( a \) from the right.
For the limit to exist at \( x = a \):
$ \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L $
Rules (Algebra of Limits)
If \( \lim_{x \to a} f(x) = l \) and \( \lim_{x \to a} g(x) = m \), then:
| Operation | Rule |
|---|---|
| Addition | \( \lim(f + g) = l + m \) |
| Subtraction | \( \lim(f – g) = l – m \) |
| Multiplication | \( \lim(f \cdot g) = lm \) |
| Division | \( \lim\dfrac{f}{g} = \dfrac{l}{m}, \, m \ne 0 \) |
| Constant Multiple | \( \lim(kf) = k \lim f \) |
Standard Limits
| Form | Value |
|---|---|
| \( \displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} \) | 1 |
| \( \displaystyle \lim_{x \to 0} \dfrac{\tan x}{x} \) | 1 |
| \( \displaystyle \lim_{x \to 0} \dfrac{1 – \cos x}{x^2} \) | \( \dfrac{1}{2} \) |
| \( \displaystyle \lim_{x \to 0} (1 + x)^{1/x} \) | \( e \) |
| \( \displaystyle \lim_{x \to 0} \dfrac{e^x – 1}{x} \) | 1 |
| \( \displaystyle \lim_{x \to 0} \dfrac{\log(1 + x)}{x} \) | 1 |
Indeterminate Forms
While evaluating limits, certain forms do not give direct answers and are called indeterminate forms.
| Indeterminate Form | Examples |
|---|---|
| \( \dfrac{0}{0} \) | \( \dfrac{x^2 – 4}{x – 2} \) |
| \( \dfrac{\infty}{\infty} \) | \( \dfrac{3x^2 + 5}{2x^2 – 7} \) |
| \( 0 \times \infty \) | \( x \log x \text{ as } x \to 0^+ \) |
| \( \infty – \infty \) | \( \tan x – \sec x \text{ as } x \to \pi/2^- \) |
| \( 1^\infty, \, 0^0, \, \infty^0 \) | \( (1 + \tfrac{1}{x})^x, \, x^x, \, (1/x)^x \) |
Methods of Evaluating Limits
- Direct substitution – if no indeterminate form arises.
- Factoring – for rational polynomials.
- Rationalization – for square roots.
- Using standard limits – for trigonometric and exponential functions.
- L’Hôpital’s Rule – for indeterminate forms \( \dfrac{0}{0} \) or \( \dfrac{\infty}{\infty} \): $ \lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f'(x)}{g'(x)} $
Example
Find \( \displaystyle \lim_{x \to 2} \dfrac{x^2 – 4}{x – 2} \).
▶️ Answer / Explanation
Direct substitution gives \( \dfrac{0}{0} \) → indeterminate.
Factorize numerator: \( \dfrac{(x – 2)(x + 2)}{x – 2} \Rightarrow x + 2 \)
Now, \( \lim_{x \to 2} (x + 2) = 4 \).
Answer: 4
Example
Evaluate \( \displaystyle \lim_{x \to 0} \dfrac{\sin 3x}{x} \).
▶️ Answer / Explanation
Using standard limit \( \lim_{x \to 0} \dfrac{\sin x}{x} = 1 \).
Let \( 3x = t \Rightarrow x = t/3 \). Then, \( \dfrac{\sin 3x}{x} = 3\dfrac{\sin t}{t} \).
\( \Rightarrow \lim_{x \to 0} \dfrac{\sin 3x}{x} = 3 \times 1 = 3 \).
Answer: 3
Example
Evaluate \( \displaystyle \lim_{x \to 0} \dfrac{(1 + 2x)^{3/x} – e^6}{x} \).
▶️ Answer / Explanation
Let \( L = \lim_{x \to 0} (1 + 2x)^{3/x} \).
As \( x \to 0 \), \( (1 + 2x)^{1/x} \to e^2 \Rightarrow L = e^{6} \).
Now apply derivative form (since numerator → 0): \( \lim_{x \to 0} \dfrac{f(x) – f(0)}{x} = f'(0) \) for \( f(x) = (1 + 2x)^{3/x} \).
Using logarithmic differentiation and limits, the result = \( 12e^6 \).
Answer: \( 12e^6 \)
Limits of Polynomials and Rational Functions
Polynomials and rational functions are among the simplest functions for evaluating limits. They can usually be solved by direct substitution or factorization.
Let \( f(x) \) be a polynomial or rational function. The limit \( \displaystyle \lim_{x \to a} f(x) \) represents the value \( f(x) \) approaches as \( x \) approaches \( a \).
Direct Substitution Method
If the function \( f(x) \) is continuous at \( x = a \), then:
$ \lim_{x \to a} f(x) = f(a) $
That is, we can simply substitute \( x = a \) in the function.
Example: \( \lim_{x \to 3} (x^2 + 2x – 5) = 9 + 6 – 5 = 10 \).
Rational Functions and Indeterminate Form \( \dfrac{0}{0} \)
If direct substitution gives \( \dfrac{0}{0} \), the function is indeterminate. We then simplify using factorization or rationalization.
Example: \( \displaystyle \lim_{x \to 2} \dfrac{x^2 – 4}{x – 2} \)
Substituting \( x = 2 \) gives \( 0/0 \). Factorize numerator: \( (x – 2)(x + 2) \Rightarrow \) cancel \( (x – 2) \).
Then \( \lim_{x \to 2} (x + 2) = 4 \).
Limits at Infinity
When \( x \to \infty \) or \( x \to -\infty \), the behavior of a polynomial or rational function depends on the degree of numerator and denominator.
Let \( f(x) = \dfrac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials.
| Case | Result | Example |
|---|---|---|
| Degree(P) < Degree(Q) | Limit = 0 | \( \displaystyle \lim_{x \to \infty} \dfrac{3x + 2}{x^3 + 5} = 0 \) |
| Degree(P) = Degree(Q) | Limit = Ratio of leading coefficients | \( \displaystyle \lim_{x \to \infty} \dfrac{2x^3 + 5}{x^3 – 1} = 2 \) |
| Degree(P) > Degree(Q) | Limit = \( \infty \) or \( -\infty \) | \( \displaystyle \lim_{x \to \infty} \dfrac{x^4 + 1}{2x^2 + 3} = \infty \) |
Method for Infinite Limits
Divide numerator and denominator by the highest power of x in the denominator:
$ \lim_{x \to \infty} \dfrac{P(x)}{Q(x)} = \lim_{x \to \infty} \dfrac{\text{coefficients of highest powers}}{\text{coefficients of highest powers}} $
Continuity and Direct Substitution Shortcut
All polynomials are continuous everywhere. Hence, for any \( a \in \mathbb{R} \):
$ \lim_{x \to a} P(x) = P(a) $
For rational functions, they are continuous except where the denominator = 0.
Example
Find \( \displaystyle \lim_{x \to 3} (x^3 – 2x^2 + 4x – 5) \).
▶️ Answer / Explanation
Since it’s a polynomial, apply direct substitution:
\( f(3) = 27 – 18 + 12 – 5 = 16 \)
Answer: 16
Example
Find \( \displaystyle \lim_{x \to 1} \dfrac{x^3 – 1}{x – 1} \).
▶️ Answer / Explanation
Substitute \( x = 1 \Rightarrow 0/0 \) form.
Factorize numerator: \( x^3 – 1 = (x – 1)(x^2 + x + 1) \).
Cancel \( (x – 1) \), then limit = \( \lim_{x \to 1} (x^2 + x + 1) = 3 \).
Answer: 3
Example
Find \( \displaystyle \lim_{x \to \infty} \dfrac{4x^3 + 2x^2 – 5}{2x^3 – 7x + 1} \).
▶️ Answer / Explanation
Divide numerator and denominator by \( x^3 \):
\( \displaystyle \lim_{x \to \infty} \dfrac{4 + \dfrac{2}{x} – \dfrac{5}{x^3}}{2 – \dfrac{7}{x^2} + \dfrac{1}{x^3}} \).
As \( x \to \infty \), all terms with \( 1/x \to 0 \).
\( \Rightarrow \dfrac{4}{2} = 2 \).
Answer: 2
Limits Involving Trigonometric and Exponential Functions
Trigonometric and exponential functions frequently appear in limit problems. Many limits in JEE are based on standard trigonometric and exponential limits combined with algebraic manipulation.
Standard Trigonometric Limits
These are fundamental and must be memorized:
| Limit Form | Value |
|---|---|
| \( \displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} \) | 1 |
| \( \displaystyle \lim_{x \to 0} \dfrac{\tan x}{x} \) | 1 |
| \( \displaystyle \lim_{x \to 0} \dfrac{1 – \cos x}{x^2} \) | \( \dfrac{1}{2} \) |
| \( \displaystyle \lim_{x \to 0} \dfrac{\sin ax}{x} \) | a |
| \( \displaystyle \lim_{x \to 0} \dfrac{\tan ax}{bx} \) | \( \dfrac{a}{b} \) |
Standard Exponential and Logarithmic Limits
| Limit Form | Value |
|---|---|
| \( \displaystyle \lim_{x \to 0} \dfrac{e^x – 1}{x} \) | 1 |
| \( \displaystyle \lim_{x \to 0} \dfrac{\log(1 + x)}{x} \) | 1 |
| \( \displaystyle \lim_{x \to \infty} \left(1 + \dfrac{1}{x}\right)^x \) | \( e \) |
| \( \displaystyle \lim_{x \to 0} (1 + ax)^{1/x} \) | \( e^a \) |
Important Combined Forms
- \( \displaystyle \lim_{x \to 0} \dfrac{\sin(ax)}{\sin(bx)} = \dfrac{a}{b} \)
- \( \displaystyle \lim_{x \to 0} \dfrac{\tan(ax)}{\sin(bx)} = \dfrac{a}{b} \)
- \( \displaystyle \lim_{x \to 0} \dfrac{1 – \cos(ax)}{x^2} = \dfrac{a^2}{2} \)
- \( \displaystyle \lim_{x \to 0} \dfrac{e^{kx} – 1}{x} = k \)
- \( \displaystyle \lim_{x \to 0} \dfrac{\log(1 + ax)}{x} = a \)
Special Angles — Conversion to Standard Form
Whenever trigonometric expressions involve coefficients with \( a \), \( b \), or higher powers, we convert them into standard limits by multiplying and dividing by the same coefficient.
Example:
$ \lim_{x \to 0} \dfrac{\sin(3x)}{x} = 3 \lim_{x \to 0} \dfrac{\sin(3x)}{3x} = 3(1) = 3 $
Indeterminate Trigonometric Forms
Common indeterminate trigonometric forms include:
- \( \dfrac{0}{0} \) → simplify using standard limits or L’Hôpital’s Rule
- \( \infty – \infty \) → use trigonometric identities
- \( 1^\infty \) → use exponential substitution
Tricks to Simplify Trigonometric Limits
- Convert everything into sine and cosine if mixed (e.g., tan, cot).
- Use standard identities:
- \( 1 – \cos x = 2\sin^2(x/2) \)
- \( \sin 2x = 2\sin x \cos x \)
- \( \tan x \approx x \) when \( x \to 0 \)
- For composite arguments like \( \sin(ax + b) \), use substitution: let \( t = ax + b \), then \( t \to b \) as \( x \to 0 \).
Example
Evaluate \( \displaystyle \lim_{x \to 0} \dfrac{\sin(5x)}{x} \).
▶️ Answer / Explanation
Multiply and divide by 5:
\( \dfrac{\sin(5x)}{x} = 5 \dfrac{\sin(5x)}{5x} \).
\( \lim_{x \to 0} \dfrac{\sin(5x)}{5x} = 1 \Rightarrow \) limit = 5.
Answer: 5
Example
Evaluate \( \displaystyle \lim_{x \to 0} \dfrac{\sin(3x)}{\sin(5x)} \).
▶️ Answer / Explanation
Write both in standard form:
\( \dfrac{\sin(3x)}{\sin(5x)} = \dfrac{\dfrac{\sin(3x)}{x}}{\dfrac{\sin(5x)}{x}} \)
\( \lim_{x \to 0} \dfrac{\sin(3x)}{\sin(5x)} = \dfrac{3}{5} \).
Answer: \( \dfrac{3}{5} \)
Example
Find \( \displaystyle \lim_{x \to 0} \dfrac{(1 + 3x)^{1/x} – e^3}{x} \).
▶️ Answer / Explanation
Let \( f(x) = (1 + 3x)^{1/x} \). As \( x \to 0 \), \( f(x) \to e^3 \).
This is a \( \dfrac{0}{0} \) form, so differentiate \( f(x) \) using logarithmic differentiation.
\( \ln f = \dfrac{\ln(1 + 3x)}{x} \Rightarrow \dfrac{f'(x)}{f(x)} = \dfrac{3x – 3(1 + 3x)\ln(1 + 3x)}{x^2(1 + 3x)} \)
After simplification and applying limit \( x \to 0 \), result = \( 9e^3 / 2 \).
Answer: \( \dfrac{9e^3}{2} \)
Limits Involving Indeterminate Forms (0/0, ∞/∞, 1^∞, 0⁰, ∞⁰)
While evaluating limits, sometimes direct substitution gives an undefined or ambiguous result. Such cases are called indeterminate forms.
These forms cannot be directly evaluated — we must simplify or transform them using algebraic methods or L’Hôpital’s Rule.
Common Indeterminate Forms
| Type | Example |
|---|---|
| \( \dfrac{0}{0} \) | \( \dfrac{x^2 – 1}{x – 1} \) |
| \( \dfrac{\infty}{\infty} \) | \( \dfrac{3x^2 + 1}{2x^2 – 5} \) |
| \( 0 \times \infty \) | \( x \log x \text{ as } x \to 0^+ \) |
| \( \infty – \infty \) | \( \tan x – \sec x \text{ as } x \to \pi/2^- \) |
| \( 1^\infty \) | \( (1 + x)^{1/x} \) |
| \( 0^0 \) | \( x^x \text{ as } x \to 0^+ \) |
| \( \infty^0 \) | \( (x)^{1/x} \text{ as } x \to \infty \) |
Algebraic Methods to Resolve Indeterminate Forms
- Factorization — for \( \dfrac{0}{0} \) forms.
- Rationalization — for expressions with square roots.
- Conjugate multiplication — useful in \( \sqrt{x + 1} – \sqrt{x} \) type limits.
- Multiplying and dividing by \( x \) — for trigonometric and exponential cases.
L’Hôpital’s Rule
If \( \displaystyle \lim_{x \to a} \dfrac{f(x)}{g(x)} \) gives \( \dfrac{0}{0} \) or \( \dfrac{\infty}{\infty} \), then (if \( f'(x) \) and \( g'(x) \) exist near \( a \)):
$ \lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f'(x)}{g'(x)} $
If the new limit again gives an indeterminate form, L’Hôpital’s Rule can be applied repeatedly.
Special Exponential and Logarithmic Indeterminate Forms
- For \( 1^\infty \) type: Convert \( y = [f(x)]^{g(x)} \Rightarrow \ln y = g(x)\ln f(x) \), then evaluate using standard limits.
- For \( 0^0 \) and \( \infty^0 \): Similar approach — take logarithm and simplify to form \( 0 \times \infty \) or \( \dfrac{0}{0} \).
Common Transformations
| Indeterminate Form | Transformation |
|---|---|
| \( 0 \times \infty \) | Convert to \( \dfrac{0}{1/∞} \) or \( \dfrac{∞}{1/0} \) |
| \( 1^\infty \) | Take \( \ln y = g(x)\ln f(x) \) |
| \( 0^0, \infty^0 \) | Take log and rewrite as exponential limit |
Special Indeterminate Examples
- \( \displaystyle \lim_{x \to 0} x^x = 1 \) (since \( \ln y = x \ln x \to 0 \))
- \( \displaystyle \lim_{x \to \infty} \left(1 + \dfrac{a}{x}\right)^x = e^a \)
- \( \displaystyle \lim_{x \to 0} (1 + \tan x)^{1/x} = e \)
Key Takeaways for JEE
- Identify the type of indeterminate form first before applying any rule.
- For \( 0/0 \) or \( ∞/∞ \) → simplify or use L’Hôpital’s Rule.
- For \( 1^\infty, 0^0, ∞^0 \) → take logarithm and use exponential limits.
- Always check existence of derivatives when using L’Hôpital’s Rule.
- Memorize key exponential transformations — these are frequent in JEE.
Common Mistakes to Avoid
- Applying L’Hôpital’s Rule to forms other than \( 0/0 \) or \( ∞/∞ \).
- Forgetting to exponentiate after taking log in exponential forms.
- Not simplifying expressions before applying derivative rules.
- Ignoring the domain where the function is valid (especially for \( \log \) or \( \sqrt{} \)).
Example
Find \( \displaystyle \lim_{x \to 2} \dfrac{x^2 – 4}{x – 2} \).
▶️ Answer / Explanation
This is \( 0/0 \) form.
Factorize: \( x^2 – 4 = (x – 2)(x + 2) \).
Cancel \( (x – 2) \), and substitute \( x = 2 \): \( f(x) = x + 2 = 4 \).
Answer: 4
Example
Evaluate \( \displaystyle \lim_{x \to 0} \dfrac{\sin x – x}{x^3} \).
▶️ Answer / Explanation
Direct substitution → \( 0/0 \).
Apply L’Hôpital’s Rule repeatedly:
\( \lim \dfrac{\sin x – x}{x^3} = \lim \dfrac{\cos x – 1}{3x^2} = \lim \dfrac{-\sin x}{6x} = -\dfrac{1}{6} \).
Answer: \( -\dfrac{1}{6} \)
Example
Find \( \displaystyle \lim_{x \to 0} (1 + 2x)^{3/x} \).
▶️ Answer / Explanation
This is of the form \( 1^\infty \).
Let \( y = (1 + 2x)^{3/x} \). Take \( \ln y = \dfrac{3}{x}\ln(1 + 2x) \).
Now, \( \lim_{x \to 0} \ln y = 3 \lim_{x \to 0} \dfrac{\ln(1 + 2x)}{x} = 3(2) = 6 \).
Therefore, \( \ln y = 6 \Rightarrow y = e^6 \).
Answer: \( e^6 \)
Notes and Study Materials
- Concepts of Limit, Continuity & Differentiability
- Concepts of Limit
- Concepts of Tangent and Normal
- Concepts of Maxima and Minima
- Concepts of Rolle’s and Lagrange’s Mean Value Theorems
- Concepts of Monotonocity
- Concepts of Logarithmic Function
- Limit, Continuity & Differentiability Master File
- Limit, Continuity & Differentiability Revision Notes
- Limit, Continuity & Differentiability Formulae
- Limit, Continuity & Differentiability Reference Book
- Limit, Continuity & Differentiability Past Many Years Questions and Answer
Examples and Exercise
- Practice Paper 1
- Practice Paper 2
- Practice Paper 3
- Practice Paper 4
- Practice Paper 5
- Practice Paper 6
- Practice Paper 7
- Practice Paper 8
IIT JEE (Main) Mathematics ,”Limit, Continuity & Differentiability” Notes ,Test Papers, Sample Papers, Past Years Papers , NCERT , S. L. Loney and Hall & Knight Solutions and Help from Ex- IITian
About this unit
Real – valued functions, algebra of functions, polynomials, rational, trigonometric, logarithmic and exponential functions, inverse functions.Graphs of simple functions.Limits, continuity, and differentiability. Differentiation of the sum, difference, product, and quotient of two functions. Differentiation of trigonometric, inverse trigonometric, logarithmic, exponential, composite and implicit functions; derivatives of order up to two. Rolle’s and Lagrange’s Mean Value Theorems. Applications of derivatives: Rate of change of quantities, monotonic – increasing and decreasing functions, Maxima, and minima of functions of one variable, tangents, and normals.
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