Continuity and Differentiability: CBSE Class 12 Maths NCERT Guide

Welcome to the fascinating world of Calculus! This chapter, Continuity and Differentiability, lays the fundamental groundwork for understanding how functions behave. Think of it as learning the grammar before you can write powerful essays in mathematics.

So, what does it mean for a function to be 'continuous'? Intuitively, it means you can draw its graph without lifting your pen from the paper. There are no sudden jumps, gaps, or holes. 'Differentiability' is a step further – it implies the graph is not only continuous but also 'smooth', without any sharp corners or cusps.

Why is this important? These concepts are the bedrock of differential calculus, which helps us study rates of change. This is crucial in physics (velocity, acceleration), economics (marginal cost), engineering, and many other fields. In this chapter, you will master the precise definitions of continuity and differentiability, learn how to test functions, and understand the vital relationship between these two key properties.

What is Continuity? The Formal Definition

The intuitive idea of drawing a graph without lifting your pen is a great start, but in mathematics, we need a precise definition. A function f(x) is said to be continuous at a point x = c in its domain if the following three conditions are met:

f(c) is defined: The function must have a value at the point c. You can't have a hole where the point should be.
The limit of the function as x approaches c exists: This means the function must approach the same value from both the left and the right side of c. We write this as:

lim(x→c⁻) f(x) = lim(x→c⁺) f(x)
Here, lim(x→c⁻) f(x) is the Left-Hand Limit (LHL) and lim(x→c⁺) f(x) is the Right-Hand Limit (RHL).

The limit equals the function's value: The value the function approaches must be the same as the function's actual value at that point.

lim(x→c) f(x) = f(c)

If any one of these three conditions fails, the function is discontinuous at x = c. For example, a simple function like f(x) = x² is continuous everywhere because for any point c, lim(x→c) x² = c² and f(c) = c².

Worked Examples: Checking for Continuity

Example 1: Check the continuity of the function f(x) = 5x - 3 at x = 1. Step 1: Check if f(1) is defined. f(1) = 5(1) - 3 = 2. The function is defined at x=1. Step 2: Calculate the Left-Hand Limit (LHL). LHL = lim(x→1⁻) f(x) = lim(h→0) f(1-h) = lim(h→0) [5(1-h) - 3] = lim(h→0) [5 - 5h - 3] = lim(h→0) [2 - 5h] = 2. Step 3: Calculate the Right-Hand Limit (RHL). RHL = lim(x→1⁺) f(x) = lim(h→0) f(1+h) = lim(h→0) [5(1+h) - 3] = lim(h→0) [5 + 5h - 3] = lim(h→0) [2 + 5h] = 2. Step 4: Compare LHL, RHL and f(1). We see that LHL = RHL = 2, so lim(x→1) f(x) = 2. Also, f(1) = 2. Final Answer: Since lim(x→1) f(x) = f(1), the function is continuous at x = 1.
Example 2: Discuss the continuity of the function f(x) at x = 2, where f(x) = {x + 1, if x ≤ 2; x - 1, if x > 2}. Step 1: Check if f(2) is defined. For x=2, we use the first definition: f(2) = 2 + 1 = 3. The function is defined. Step 2: Calculate the LHL at x=2. LHL = lim(x→2⁻) f(x). Since x is approaching 2 from the left, x < 2, so we use f(x) = x + 1. LHL = lim(x→2⁻) (x+1) = 2 + 1 = 3. Step 3: Calculate the RHL at x=2. RHL = lim(x→2⁺) f(x). Since x is approaching 2 from the right, x > 2, so we use f(x) = x - 1. RHL = lim(x→2⁺) (x-1) = 2 - 1 = 1. Step 4: Compare LHL and RHL. Here, LHL = 3 and RHL = 1. Since LHL ≠ RHL, the limit lim(x→2) f(x) does not exist. Final Answer: As the limit does not exist (condition 2 fails), the function f(x) is discontinuous at x = 2.

Understanding Differentiability

Differentiability is a stricter, stronger condition than continuity. If a function is differentiable at a point, it is guaranteed to be continuous at that point. However, the reverse is not always true; a continuous function might not be differentiable.

Intuitively, differentiability means the function's graph is 'smooth' at a point. It doesn't have any sharp corners, cusps, or vertical tangents. The classic example of a continuous but non-differentiable function is f(x) = |x|, which has a sharp corner at x=0.

The formal definition of a derivative gives us the tool to test for differentiability. A function f(x) is differentiable at a point x = c if the following limit exists:
f'(c) = lim(h→0) [f(c+h) - f(c)] / h

For this limit to exist, the limit from the left (Left-Hand Derivative, LHD) must equal the limit from the right (Right-Hand Derivative, RHD).

LHD: lim(h→0⁻) [f(c+h) - f(c)] / h
RHD: lim(h→0⁺) [f(c+h) - f(c)] / h

If LHD = RHD, the function is differentiable at x = c. If they are not equal, the function is not differentiable at x=c.

Worked Example: Checking for Differentiability

Example: Check if the function f(x) = |x| is differentiable at x=0. Step 1: Check for continuity (a necessary first step). LHL = lim(x→0⁻) |x| = lim(x→0⁻) -x = 0. RHL = lim(x→0⁺) |x| = lim(x→0⁺) x = 0. And f(0) = |0| = 0. Since LHL = RHL = f(0), the function is continuous at x=0. Now we can check for differentiability. Step 2: Calculate the Left-Hand Derivative (LHD) at x=0. LHD = lim(h→0⁻) [f(0+h) - f(0)] / h = lim(h→0⁻) [|h| - 0] / h. Since h is approaching 0 from the left, h < 0, so |h| = -h. LHD = lim(h→0⁻) -h/h = lim(h→0⁻) -1 = -1. Step 3: Calculate the Right-Hand Derivative (RHD) at x=0. RHD = lim(h→0⁺) [f(0+h) - f(0)] / h = lim(h→0⁺) [|h| - 0] / h. Since h is approaching 0 from the right, h > 0, so |h| = h. RHD = lim(h→0⁺) h/h = lim(h→0⁺) 1 = 1. Step 4: Compare LHD and RHD. We have LHD = -1 and RHD = 1. Final Answer: Since LHD ≠ RHD, the function f(x) = |x| is not differentiable at x=0.

Important Points & Exam Tips

1. Differentiability implies Continuity: This is a crucial one-mark concept. If a function is differentiable at a point, it must be continuous there. The converse is false. f(x) = |x| at x=0 is the perfect counterexample to remember.

2. Break-Points are Critical: For piecewise functions, the points where the function definition changes are the most likely candidates for discontinuity or non-differentiability. Always test these points thoroughly.

3. Standard Function Properties: Know the properties of standard functions.

Polynomials, e^x, sin(x), cos(x) are continuous and differentiable everywhere.
log(x) is continuous and differentiable for x > 0.
Greatest Integer Function [x] is discontinuous at all integer points.

4. Quick Check for Differentiability: A useful shortcut for piecewise functions: First, check for continuity. If it's continuous, then differentiate each piece of the function separately. If the derivatives of the pieces are equal at the break-point, the function is likely differentiable. This is a quick check, but for board exams, always use the LHD/RHD first principles method for full marks.

Practice Questions with Solutions

Q: Is the function defined by f(x) = {x + 5, if x ≤ 1; x - 5, if x > 1} continuous at x = 1? A: Step 1: Find the value of the function at x=1. According to the definition, for x ≤ 1, f(x) = x + 5. So, f(1) = 1 + 5 = 6. Step 2: Calculate the Left-Hand Limit (LHL). LHL = lim(x→1⁻) f(x) = lim(x→1⁻) (x + 5) = 1 + 5 = 6. Step 3: Calculate the Right-Hand Limit (RHL). RHL = lim(x→1⁺) f(x) = lim(x→1⁺) (x - 5) = 1 - 5 = -4. Step 4: Compare the limits. Here, LHL ≠ RHL (6 ≠ -4). Therefore, the limit lim(x→1) f(x) does not exist. Final answer: The function is not continuous at x = 1.
Q: Find all points of discontinuity of the function f(x) = |x| - |x+1|. A: Step 1: Identify the functions involved. Both |x| and |x+1| are continuous functions for all real numbers. Step 2: Recall the algebra of continuous functions. The difference of two continuous functions is also a continuous function. Step 3: Apply the principle. Since g(x) = |x| and h(x) = |x+1| are continuous everywhere, their difference f(x) = g(x) - h(x) must also be continuous everywhere. Final answer: There are no points of discontinuity for the function f(x) = |x| - |x+1|.
Q: Prove that the greatest integer function f(x) = [x] is discontinuous at x=2. A: Step 1: Find the value of the function at x=2. f(2) = [2] = 2. Step 2: Calculate the LHL. LHL = lim(x→2⁻) [x]. For x slightly less than 2 (e.g., 1.999), the greatest integer is 1. So, LHL = 1. Step 3: Calculate the RHL. RHL = lim(x→2⁺) [x]. For x slightly greater than 2 (e.g., 2.001), the greatest integer is 2. So, RHL = 2. Step 4: Compare the limits and the function value. We have LHL = 1 and RHL = 2. Since LHL ≠ RHL, the limit at x=2 does not exist. Final answer: The function f(x) = [x] is discontinuous at x=2.
Q: Show that the function f(x) = x³ is differentiable at x=1 and find its derivative. A: Step 1: Use the first principle of derivatives: f'(c) = lim(h→0) [f(c+h) - f(c)] / h. Here, c=1. Step 2: Substitute the function and the point into the formula. f'(1) = lim(h→0) [f(1+h) - f(1)] / h = lim(h→0) [(1+h)³ - 1³] / h. Step 3: Expand the numerator. (1+h)³ = 1³ + 3(1)²(h) + 3(1)(h)² + h³ = 1 + 3h + 3h² + h³. The expression becomes lim(h→0) [1 + 3h + 3h² + h³ - 1] / h. Step 4: Simplify the expression. lim(h→0) [3h + 3h² + h³] / h = lim(h→0) h(3 + 3h + h²) / h = lim(h→0) (3 + 3h + h²). Step 5: Evaluate the limit by substituting h=0. f'(1) = 3 + 3(0) + (0)² = 3. Final answer: Since the limit exists and is a finite number (3), the function is differentiable at x=1, and its derivative is 3.

Frequently Asked Questions

What is the main difference between continuity and differentiability?

Continuity means a function's graph has no breaks, jumps, or holes. Differentiability is a stricter condition; it means the graph is continuous AND smooth, with no sharp corners or cusps. Every differentiable function is continuous, but not every continuous function is differentiable.

Are all polynomial functions continuous and differentiable?

Yes. All polynomial functions (like `f(x) = ax^n + bx^(n-1) + ...`) are continuous and differentiable for all real numbers. Their graphs are always smooth, unbroken curves.

How do I test if a function is continuous at a point x=c?

You must verify three conditions. First, `f(c)` must be defined. Second, the limit as x approaches c must exist (meaning LHL = RHL). Third, the limit must equal the function's value (`lim(x→c) f(x) = f(c)`).

Why is the function f(x) = |x| not differentiable at x=0?

Geometrically, it has a sharp corner at x=0, so there's no unique tangent. Mathematically, the Left-Hand Derivative (LHD) at x=0 is -1, while the Right-Hand Derivative (RHD) is +1. Since LHD ≠ RHD, the derivative does not exist.