Continuity & Differentiability

Class 12 Mathematics • Chapter 5 • NCERT Part I • MyAiSchool

5.1 Introduction

In Class 11 we introduced the idea of a limit of a function at a point. We informally examined derivatives of polynomial and trigonometric functions. In this chapter we sharpen those ideas: we study continuity, differentiability, and their interrelations; introduce many new rules (chain rule, derivatives of implicit and inverse-trig functions, exponential/logarithmic functions, logarithmic differentiation, parametric forms) and extend to second-order derivatives.

5.2 Continuity

If \(f\) is continuous at every point of its domain, we say \(f\) is a continuous function. If \(f\) is not continuous at \(c\) we say \(f\) has a discontinuity at \(c\) and \(c\) is a point of discontinuity.

Example 1 — Checking continuity of \(f(x)=2x+3\) at \(x=1\)

\(\lim_{x\to 1}(2x+3)=2(1)+3=5\) and \(f(1)=5\). Since the limit equals the value, \(f\) is continuous at \(x=1\).

Example 2 — Step function

Let \(f(x)=\begin{cases}1,& x\le 0\\ 2,& x>0\end{cases}\). Left-hand limit at 0 is 1, right-hand limit is 2. They are unequal, so \(\lim_{x\to 0}f(x)\) does not exist, and \(f\) is discontinuous at \(0\).

Example 3 — Constant function

Any constant function \(f(x)=k\) is continuous everywhere because \(\lim_{x\to c}k=k=f(c)\).

Example 4 — Identity and polynomial functions

For \(f(x)=x\), \(\lim_{x\to c}x=c=f(c)\), so identity is continuous. By repeated products and sums, every polynomial \(p(x)=a_0+a_1x+\cdots+a_nx^n\) is continuous on \(\mathbb R\).

Example 5 — Modulus function

\(f(x)=|x|=\begin{cases}-x,&x<0\\ x,& x\ge 0\end{cases}\). At \(x=0\): LHL \(=\lim_{x\to 0^-}(-x)=0\), RHL \(=\lim_{x\to 0^+}x=0\), and \(f(0)=0\). Hence \(|x|\) is continuous at \(0\); in fact it is continuous everywhere.

Example 6 — \(1/x\)

\(f(x)=1/x,\;x\neq 0\). For any \(c\neq 0\), \(\lim_{x\to c}1/x=1/c=f(c)\). So \(f\) is continuous on its domain. Note \(0\) is not in the domain; we do not ask about continuity at \(0\).

Example 7 — Piecewise around a point

Let \(f(x)=\begin{cases}x+2,&x\le 1\\ x-2,&x>1\end{cases}\). LHL at 1 is \(3\), RHL is \(-1\); unequal, so discontinuous at 1.

5.2.1 Algebra of Continuous Functions

Proof sketch of (i). \(\lim_{x\to c}(f+g)(x)=\lim_{x\to c}f(x)+\lim_{x\to c}g(x)=f(c)+g(c)=(f+g)(c).\) Parts (ii)–(iv) follow from the corresponding limit laws.

Composition of continuous functions

Theorem. If \(g\) is continuous at \(c\) and \(f\) is continuous at \(g(c)\), then \(f\circ g\) is continuous at \(c\).

Continuity of trigonometric functions

\(\sin x\) and \(\cos x\) are continuous everywhere. Recall \(|\sin x-\sin c|\le |x-c|\); as \(x\to c\) the right side \(\to 0\). Hence \(\lim_{x\to c}\sin x=\sin c\). Similarly for \(\cos\). Then \(\tan x=\sin x/\cos x\) is continuous wherever \(\cos x\neq 0\), i.e. \(x\neq (2n+1)\pi/2\).

Worked Examples