Core Concept Reading Time: 15 min JEE Essential

Continuity and Differentiability: The Critical Relationship for JEE

Understanding why differentiability implies continuity, but continuity doesn't guarantee differentiability - with proofs, examples, and JEE strategies.

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Why This Relationship Matters in JEE

The relationship between continuity and differentiability is one of the most frequently tested concepts in JEE Mathematics. Understanding this relationship helps you:

Solve 2-3 questions in every JEE paper
Avoid common traps in calculus problems
Build strong foundation for advanced topics
Save precious time during examination

The Fundamental Relationship

Differentiable at a point → Continuous at that point

But Continuous at a point ↛ Differentiable at that point

🎯 Quick Navigation

Definitions Mathematical Proof Counterexamples JEE Application

1. Understanding the Definitions

Continuity at a Point

A function $f(x)$ is continuous at $x = a$ if:

Three conditions must be satisfied:

$f(a)$ exists (function is defined at $x = a$)
$\lim_{x \to a} f(x) = f(a)$

In simple terms: No breaks, jumps, or holes at $x = a$

Differentiability at a Point

A function $f(x)$ is differentiable at $x = a$ if:

The derivative exists:

$$ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \quad \text{exists} $$

Left-hand and right-hand derivatives must be equal:

$$ Lf'(a) = Rf'(a) $$

In simple terms: Smooth curve with a well-defined tangent at $x = a$

2. The Mathematical Proof: Differentiability → Continuity

Theorem Statement

If a function is differentiable at a point, then it is continuous at that point.

$$ \text{Differentiable at } x = a \Rightarrow \text{Continuous at } x = a $$

Step-by-Step Proof

Step 1: Assume differentiability

Assume $f(x)$ is differentiable at $x = a$, so:

$$ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \quad \text{exists} $$

Step 2: Express f(a+h) - f(a)

We can write:

$$ f(a+h) - f(a) = \frac{f(a+h) - f(a)}{h} \cdot h $$

Step 3: Take limit as h → 0

$$ \lim_{h \to 0} [f(a+h) - f(a)] = \lim_{h \to 0} \left[ \frac{f(a+h) - f(a)}{h} \cdot h \right] $$

Using limit properties:

$$ = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \cdot \lim_{h \to 0} h $$

$$ = f'(a) \cdot 0 = 0 $$

Step 4: Conclude continuity

Since $\lim_{h \to 0} [f(a+h) - f(a)] = 0$, we have:

$$ \lim_{h \to 0} f(a+h) = f(a) $$

Which means $f(x)$ is continuous at $x = a$.

✓ Proof Complete

We have shown that differentiability at a point implies continuity at that point.

3. Counterexamples: Continuous but Not Differentiable

⚠️ Important Note

The converse is not true. A function can be continuous at a point but not differentiable there.

Example 1: Absolute Value Function

Consider $f(x) = |x|$ at $x = 0$

Continuity check:

$f(0) = 0$ ✓
$\lim_{x \to 0} |x| = 0$ ✓
$\lim_{x \to 0} f(x) = f(0)$ ✓

Conclusion: Continuous at $x = 0$

Differentiability check:

Left-hand derivative:

$$ Lf'(0) = \lim_{h \to 0^-} \frac{|0+h| - |0|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1 $$

Right-hand derivative:

$$ Rf'(0) = \lim_{h \to 0^+} \frac{|0+h| - |0|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 $$

Conclusion: $Lf'(0) \neq Rf'(0)$, so not differentiable at $x = 0$

Example 2: Cube Root Function

Consider $f(x) = \sqrt[3]{x}$ at $x = 0$

Continuity check:

$f(0) = 0$ ✓
$\lim_{x \to 0} \sqrt[3]{x} = 0$ ✓
$\lim_{x \to 0} f(x) = f(0)$ ✓

Conclusion: Continuous at $x = 0$

Differentiability check:

$$ f'(0) = \lim_{h \to 0} \frac{\sqrt[3]{h} - 0}{h} = \lim_{h \to 0} \frac{1}{\sqrt[3]{h^2}} = \infty $$

Conclusion: Derivative is infinite, so not differentiable at $x = 0$

Common Cases Where Continuous ≠ Differentiable

Sharp Corners

$f(x) = |x|$, $f(x) = |x-1| + |x+1|$

Left and right derivatives differ

Vertical Tangents

$f(x) = \sqrt[3]{x}$, $f(x) = \sqrt{x}$ at $x=0$

Derivative becomes infinite

Cusps

$f(x) = x^{2/3}$

Sharp point with infinite derivatives from both sides

Discontinuity in Derivative

Piecewise functions with different formulas

Function continuous but derivative jumps

4. JEE Application & Problem Solving

JEE Problem Solving Strategy

🎯 Quick Decision Framework

When asked "Is f(x) differentiable at x=a?":

First check if f(x) is continuous at x=a
If NOT continuous → Automatically NOT differentiable
If continuous → Check left-hand and right-hand derivatives
If LHD = RHD → Differentiable
If LHD ≠ RHD → Not differentiable

JEE Main 2023 Type Question

Let $f(x) = \begin{cases} x^2 & \text{if } x \leq 1 \\ ax + b & \text{if } x > 1 \end{cases}$

Find values of a and b such that f(x) is differentiable at x = 1.

Step 1: Continuity Condition

For differentiability, first ensure continuity:

$$ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) $$

$$ 1^2 = a(1) + b = 1 $$

So: $a + b = 1$

Step 2: Differentiability Condition

Left-hand derivative:

$$ Lf'(1) = \lim_{h \to 0^-} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0^-} \frac{(1+h)^2 - 1}{h} = 2 $$

Right-hand derivative:

$$ Rf'(1) = \lim_{h \to 0^+} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0^+} \frac{a(1+h)+b - 1}{h} = a $$

For differentiability: $Lf'(1) = Rf'(1) \Rightarrow a = 2$

Step 3: Solve System

From continuity: $a + b = 1$

From differentiability: $a = 2$

So: $2 + b = 1 \Rightarrow b = -1$

Common JEE Question Types

Type 1: Direct Application

"If f(x) is differentiable at x=a, which of the following must be true?"

Answer: f(x) is continuous at x=a

Type 2: Counterexample Identification

"Which function is continuous but not differentiable?"

Answer: Functions with corners/cusps

Type 3: Parameter Finding

"Find parameters for differentiability"

Use both continuity and LHD=RHD conditions

Type 4: Graphical Analysis

"Identify points of non-differentiability from graph"

Look for corners, cusps, discontinuities

🎯 Memory Aids & Quick Recall

Relationship Summary

✓ Differentiable → Continuous

✗ Continuous → Differentiable

✗ Not Continuous → Not Differentiable

Quick Checks

Corner: |x| type functions
Cusp: $x^{2/3}$ type functions
Vertical tangent: $\sqrt[3]{x}$ type functions
Discontinuity: Jump/removable/infinite

📝 Practice Problems

Test your understanding with these JEE-style problems:

1. Prove that if f(x) is differentiable at x=2, then it must be continuous at x=2.

Hint: Use the definition of derivative

2. Show that f(x) = |x-3| is continuous but not differentiable at x=3.

Hint: Calculate left-hand and right-hand derivatives

3. Find a and b if $f(x) = \begin{cases} ax^2+b & x<2 \\ 3x-1 & x\geq2 \end{cases}$ is differentiable at x=2.

Hint: Use both continuity and differentiability conditions

Key Takeaway for JEE Success

Remember: Differentiability is a stronger condition than continuity. Every differentiable function is continuous, but not every continuous function is differentiable.