Chapter 5: Continuity and Differentiability

This chapter builds on the limit of functions to define continuity and introduces the powerful tool of derivative for finding the rate of change.

Continuity

A function is continuous at $x = c$ if the function is defined at $x = c$ and if the value of the function at $x = c$ equals the limit of the function at $x = c$. Mathematically, a function $f(x)$ is continuous at $x = c$ if: $$ \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) $$

Continuity means that the graph of the function has no break, jump, or hole at that point.

Differentiability

Let $f$ be a real function and $c$ be a point in its domain. The derivative of $f$ at $c$ is defined by: $$ f’(c) = \lim_{h \to 0} \frac{f(c + h) - f(c)}{h} $$ provided this limit exists. If $f’(c)$ exists, we say $f$ is differentiable at $c$.

Theorem: If a function $f$ is differentiable at a point $c$, then it is also continuous at that point. Note: The converse is not true. Every continuous function is not differentiable. (e.g., $f(x) = |x|$ is continuous at $x=0$ but not differentiable at $x=0$).

Chain Rule

Let $f$ be a real valued function which is a composite of two functions $u$ and $v$; i.e., $f = v \circ u$. Suppose $t = u(x)$ and if both $\frac{dt}{dx}$ and $\frac{dv}{dt}$ exist, we have $$ \frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx} $$

Competency-Based Questions

1. [CBSE 2021] Examine the continuity of the function $f(x)$ at $x = 0$, where $$ f(x) = \begin{cases} \frac{\sin 3x}{x}, & \text{if } x \neq 0 \\ 3, & \text{if } x = 0 \end{cases} $$

Solution:

For checking continuity at $x = 0$, we need to find the limit of $f(x)$ as $x \to 0$ and compare it with $f(0)$.

$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin 3x}{x}$

Multiply and divide by 3:

$= \lim_{x \to 0} 3 \frac{\sin 3x}{3x}$

Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, we have:

$\lim_{x \to 0} f(x) = 3(1) = 3$

Also, it is given that $f(0) = 3$.

Since $\lim_{x \to 0} f(x) = f(0) = 3$, the function is continuous at $x = 0$.

2. [Sample Paper 2023] Find $\frac{dy}{dx}$ if $x^2 + xy + y^2 = 100$.

Solution:

Differentiating both sides with respect to $x$:

$\frac{d}{dx}(x^2 + xy + y^2) = \frac{d}{dx}(100)$

$2x + (x \frac{dy}{dx} + y \cdot 1) + 2y \frac{dy}{dx} = 0$

$2x + y + (x + 2y)\frac{dy}{dx} = 0$

$(x + 2y)\frac{dy}{dx} = -(2x + y)$

$\frac{dy}{dx} = \frac{-(2x + y)}{x + 2y}$

3. [CBSE 2019] Differentiate $(\sin x)^{\cos x}$ with respect to $x$.

Solution:

Let $y = (\sin x)^{\cos x}$

Taking logarithm on both sides:

$\log y = \log (\sin x)^{\cos x} = \cos x \log(\sin x)$

Differentiating both sides with respect to $x$:

$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cos x \cdot \log(\sin x))$

Using product rule:

$\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \frac{d}{dx}(\log(\sin x)) + \log(\sin x) \cdot \frac{d}{dx}(\cos x)$

$\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \left( \frac{1}{\sin x} \cdot \cos x \right) + \log(\sin x) \cdot (-\sin x)$

$\frac{1}{y} \frac{dy}{dx} = \cos x \cot x - \sin x \log(\sin x)$

$\frac{dy}{dx} = y [\cos x \cot x - \sin x \log(\sin x)]$

$\frac{dy}{dx} = (\sin x)^{\cos x} [\cos x \cot x - \sin x \log(\sin x)]$