JEE Limits, Continuity, and Differentiability Notes by PW

Limits, continuity, and differentiability form the foundation of calculus in Class 12 Mathematics. Questions involving piecewise functions, greatest-integer functions, modulus, standard limits, indeterminate forms, and parameter-based conditions require careful evaluation of function behaviour near specific points. Applying the correct formula and checking one-sided conditions can help avoid common calculation errors in JEE Main and JEE Advanced.

PW supports JEE Mathematics preparation through concept-focused classes and practice resources that help students revise calculus concepts and apply them to different question types. Live and recorded classes, practice questions, previous years’ questions (PYQs), and tests support concept revision, numerical practice, and progress tracking.

Limits

Limits support the study of continuity, while continuity is a necessary condition for differentiability.

Meaning of a Limit

Consider the function: f(x) = (x − 3)/(x − 3)

For x ≠ 3, the numerator and denominator cancel, giving f(x) = 1. However, f(3) is undefined because the original expression becomes 0/0.

A limit describes the value that a function approaches as x approaches a particular point, even when the function itself is undefined at that point.

Therefore, limₓ→₃ [(x − 3)/(x − 3)] = 1

The function approaches 1 from both sides of x = 3, so the limit exists even though f(3) is undefined.

Left-Hand and Right-Hand Limits

A point can be approached from two directions.

• Right-Hand Limit (RHL): The value approached as x tends to a from values greater than a.
  RHL = limₓ→ₐ⁺ f(x) = limₕ→₀⁺ f(a + h)

• Left-Hand Limit (LHL): The value approached as x tends to a from values less than a.
  LHL = limₓ→ₐ⁻ f(x) = limₕ→₀⁺ f(a − h)

For a two-sided limit to exist, both one-sided limits must be equal and finite:

limₓ→ₐ⁻ f(x) = limₓ→ₐ⁺ f(x) = L

When this condition holds, limₓ→ₐ f(x) = L

Memory tip: Always check both one-sided limits when the function changes its expression at a point.

Greatest-Integer and Piecewise Functions

For functions involving the greatest-integer function, fractional-part function, modulus, or different formulas over separate intervals, evaluate the LHL and RHL independently.

For a point x = a, use:

• LHL: Substitute x = a − h, where h → 0⁺.

• RHL: Substitute x = a + h, where h → 0⁺.

The greatest-integer function can change abruptly at integer values, so the two sides may produce different results.

For example, for f(x) = ⌊x⌋:

limₓ→₂⁻ ⌊x⌋ = 1

limₓ→₂⁺ ⌊x⌋ = 2

Since the one-sided limits are unequal, limₓ→₂ ⌊x⌋ does not exist.

In parameter-based questions, compare the LHL and RHL and solve the resulting equation to determine the unknown parameter.

Indeterminate Forms

The seven standard indeterminate forms are:

• 0/0

• ∞/∞

• ∞ − ∞

• 0 × ∞

• 1^∞

• 0⁰

• ∞⁰

These forms arise when quantities approach particular values during a limiting process. They do not directly determine the value of a limit; further simplification is required.

For example, 1^∞ is indeterminate because the base approaches 1 while the exponent grows without bound. The final result depends on how quickly the base approaches 1 and how the exponent changes.

Algebra of Limits

Suppose: limₓ→ₐ f(x) = L and limₓ→ₐ g(x) = M

When the required limits exist, the following rules apply:

• Sum: limₓ→ₐ [f(x) + g(x)] = L + M

• Difference: limₓ→ₐ [f(x) − g(x)] = L − M

• Product: limₓ→ₐ [f(x)g(x)] = LM

• Quotient: limₓ→ₐ [f(x)/g(x)] = L/M, provided M ≠ 0

Memory tip: Limit laws can simplify sums, differences, products, and quotients. For a quotient, the limiting denominator must be nonzero.

Methods for Evaluating the 0/0 Form

When direct substitution gives 0/0, simplify the expression before evaluating the limit. Common methods include factorisation, rationalisation, standard limits, substitution, and L’Hôpital’s Rule where applicable.

Factorisation

Consider: limₓ→₂ [(x² − 4)/(x − 2)]

Direct substitution gives 0/0. Factorise the numerator: x² − 4 = (x − 2)(x + 2)

For x ≠ 2,

(x² − 4)/(x − 2) = x + 2

Therefore, limₓ→₂ [(x² − 4)/(x − 2)] = limₓ→₂ (x + 2) = 4

The common factor is cancelled for nearby values of x, not at the point where the original expression is undefined.

Rationalisation

Rationalisation is useful when an expression contains square roots. Multiply the numerator and denominator, or the expression itself, by the appropriate conjugate.

The key identity is: (a − b)(a + b) = a² − b²

For example,

limₓ→₀ [(√(x + 4) − 2)/x]

Multiply by the conjugate:

= limₓ→₀ [(√(x + 4) − 2)(√(x + 4) + 2)]/[x(√(x + 4) + 2)]

Using the difference of squares,

= limₓ→₀ [x/(x(√(x + 4) + 2))]

= limₓ→₀ [1/(√(x + 4) + 2)]

= 1/4

L’Hôpital’s Rule

L’Hôpital’s Rule is used for limits that produce the indeterminate forms 0/0 or ∞/∞, provided the relevant differentiability and other conditions are satisfied.

For a limit of the form,

limₓ→ₐ [f(x)/g(x)]

if the required conditions hold and direct substitution gives 0/0 or ∞/∞, then:

limₓ→ₐ [f(x)/g(x)] = limₓ→ₐ [f′(x)/g′(x)]

The derivative of the numerator and the derivative of the denominator are calculated separately. If the resulting limit remains indeterminate, the rule may be applied again when its conditions continue to hold.

Memory tip: First identify the indeterminate form. Do not apply L’Hôpital’s Rule automatically to every limit.

Standard Limits

Standard limits are useful for simplifying trigonometric, exponential, and logarithmic expressions.

Here, sin⁻¹x and tan⁻¹x denote inverse trigonometric functions, not reciprocals.

Standard Limits with a Constant

For constants a and b, where b ≠ 0: limₓ→₀ [sin(ax)/(bx)] = a/b

Similarly, limₓ→₀ [tan(ax)/(bx)] = a/b

These results follow by rewriting the expressions in terms of the standard limits, provided the relevant expressions are defined near zero.

Important distinction:

limₓ→₀ (sin x)/x = 1

But,

limₓ→∞ (sin x)/x = 0

The point that x approaches changes the result.

Limits at Infinity

For rational functions, compare the degrees of the numerator and denominator to evaluate limits as x approaches positive or negative infinity.

Consider:

limₓ→∞ [P(x)/Q(x)]

where P(x) and Q(x) are polynomials.

For example,

limₓ→∞ [(x + 3)/(x + 2)]

Divide the numerator and denominator by x:

= limₓ→∞ [(1 + 3/x)/(1 + 2/x)]

As x → ∞, both 3/x and 2/x approach zero. Therefore,

limₓ→∞ [(x + 3)/(x + 2)] = 1

Other methods include dividing by the highest power of x or using the substitution t = 1/x, which converts x → ∞ into t → 0⁺.

Other Indeterminate Forms

Infinity Minus Infinity

When two expressions independently approach infinity, their difference cannot be evaluated by simply subtracting the two limits. Algebraic simplification or rationalisation may be required.

For example,

limₓ→∞ [√(x² + 8x) − x]

Multiply by the conjugate:

= limₓ→∞ [(x² + 8x) − x²]/[√(x² + 8x) + x]

= limₓ→∞ [8x/(√(x² + 8x) + x)]

Divide the numerator and denominator by x:

= limₓ→∞ [8/(√(1 + 8/x) + 1)]

= 8/(1 + 1) = 4

Therefore,

limₓ→∞ [√(x² + 8x) − x] = 4

Zero Multiplied by Infinity

A product with the form 0 × ∞ is indeterminate. Convert the product into a quotient and simplify it using algebraic identities or standard limits.

For example, if a trigonometric expression contains sec x, use:

sec x = 1/cos x

The resulting quotient may be evaluated using standard limits or other applicable methods.

One Raised to Infinity

For an expression of the form [f(x)]^g(x), where f(x) → 1 and g(x) → ∞, take logarithms to evaluate the limit.

Let, y = [f(x)]^g(x)

Then, ln y = g(x) ln f(x)

If the limit of g(x) ln f(x) is L, and the required conditions hold, then: lim y = eᴸ

A common standard result is: limₓ→₀ (1 + x)^(1/x) = e

The expression is indeterminate in its original form, so its exponent and base must be considered together.

Zero Raised to Zero and Infinity Raised to Zero

For expressions of the form [f(x)]^g(x), logarithms can help convert the limit into a product or quotient.

Let, y = [f(x)]^g(x)

Taking natural logarithms gives: ln y = g(x) ln f(x)

Evaluate the limit of the right-hand side and then exponentiate the result, provided the expression is defined and the required limits exist.

For example, limₕ→₀⁺ hʰ

Taking logarithms,

ln y = h ln h

Since limₕ→₀⁺ h ln h = 0,

limₕ→₀⁺ ln y = 0

Therefore,

limₕ→₀⁺ hʰ = e⁰ = 1

Sandwich Theorem

The Sandwich Theorem, also called the Squeeze Theorem, helps evaluate a limit by bounding a function between two other functions.

If, f(x) ≤ h(x) ≤ g(x)

and,

limₓ→ₐ f(x) = limₓ→ₐ g(x) = L

then,

limₓ→ₐ h(x) = L

Application to Fractional-Part and Greatest-Integer Functions

The fractional-part function satisfies:

0 ≤ {2x} < 1

For x → ∞, divide by 2x:

0 ≤ {2x}/(2x) < 1/(2x)

Both outer expressions approach zero. Hence,

limₓ→∞ [{2x}/(2x)] = 0

For the greatest-integer function, use:

2x − 1 < ⌊2x⌋ ≤ 2x

For x → ∞, divide by 2x:

1 − 1/(2x) < ⌊2x⌋/(2x) ≤ 1

Both bounds approach 1. Therefore,

limₓ→∞ [⌊2x⌋/(2x)] = 1

The Sandwich Theorem is particularly useful when the function is bounded but difficult to evaluate directly.

Continuity

A function f(x) is continuous at x = c if its left-hand limit and right-hand limit are equal to the function value.

The condition is: limₓ→c⁻ f(x) = limₓ→c⁺ f(x) = f(c)

Equivalently, limₓ→c f(x) = f(c)

For continuity, the two-sided limit must exist, the function value must be defined, and both must be equal.

A function is discontinuous at c if:

• The LHL and RHL are unequal.

• The two-sided limit does not exist.

• The function is undefined at c.

• The limit exists but differs from f(c).

Continuity on a Closed Interval

For a function to be continuous on [a, b]:

• It must be right-continuous at a.

• It must be left-continuous at b.

• It must be continuous at every point in (a, b).

Polynomials are continuous for all real values. Trigonometric, exponential, and logarithmic functions are continuous throughout their respective domains.

Greatest-Integer and Fractional-Part Functions

The greatest-integer function ⌊x⌋ and fractional-part function {x} are discontinuous at integer values of x.

For composite functions such as ⌊f(x)⌋ or {f(x)}, possible discontinuities occur where:

f(x) ∈ ℤ

where ℤ represents the set of integers.

For the signum function sgn(f(x)), possible discontinuities occur at points where: f(x) = 0

These candidate points are often called doubtful points of discontinuity (DPDs). Each point must be checked using the relevant one-sided limits and function value.

Differentiability

A function is differentiable at x = c if its derivative exists as a finite value at that point.

The left-hand derivative (LHD) is: LHD = limₕ→₀⁻ [f(c + h) − f(c)]/h

The right-hand derivative (RHD) is: RHD = limₕ→₀⁺ [f(c + h) − f(c)]/h

The differentiability condition is: LHD = RHD = a finite value

A function must be continuous at a point to be differentiable there. However, continuity alone does not guarantee differentiability.

Therefore, Differentiability ⇒ Continuity

But, Continuity ⇏ Differentiability

Example: Modulus Function

Consider: f(x) = |x|

At x = 0, the function is continuous because: limₓ→₀ |x| = 0 = f(0)

However, its one-sided derivatives differ:

RHD = 1

LHD = −1

Since LHD ≠ RHD, f(x) = |x| is not differentiable at x = 0.

Modulus Functions of the Form |u(x)|

For functions involving |u(x)|, first solve: u(x) = 0

These points are candidates for changes in the expression and must be checked for differentiability.

However, not every such point is necessarily nondifferentiable.

For example,

f(x) = x|x|

For x ≥ 0, f(x) = x².

For x < 0, f(x) = −x².

At x = 0,

RHD = limₕ→₀⁺ [(h|h|) − 0]/h = limₕ→₀⁺ |h| = 0

LHD = limₕ→₀⁻ [(h|h|) − 0]/h = limₕ→₀⁻ |h| = 0

Since both one-sided derivatives are equal and finite, f(x) = x|x| is differentiable at zero.

Maximum and Minimum Functions

For functions involving maximum or minimum operations, use the following steps:

1. Find the intersection points of the component functions.

2. Identify which function gives the maximum value in each interval.

3. Identify which function gives the minimum value in each interval.

4. Check differentiability at points where the selected expression changes.

If the left-hand and right-hand derivatives differ at a branch-changing point, the function is not differentiable there. A corner may indicate nondifferentiability, but the one-sided derivative test provides the mathematical condition.

Limits, Continuity, and Differentiability form an important part of Class 12 Mathematics and JEE preparation. Regular practice is essential to strengthen these concepts and apply formulas correctly. PW supports your JEE Main and JEE Advanced preparation with concept-focused classes, live and recorded learning, practice questions, previous years’ questions (PYQs), and tests. These resources help you revise key concepts, practise different question patterns, and assess your understanding of calculus.

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