Continuity

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Definition: A function $f$ is continuous at $x = a$ if: $$\lim_{x \to a} f(x) = f(a)$$

Equivalently: $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$.

Types of Discontinuity:

1. Removable: Limit exists but not equal to $f(a)$ (or $f(a)$ undefined)
   • Example: $f(x) = \frac{x^2-1}{x-1}$ at $x=1$ (limit is 2, but define $f(1)=3$ removes it)
2. Jump (Finite discontinuity): Left and right limits exist but are different
   • Example: signum function $\text{sgn}(x)$
3. Infinite discontinuity: One or both one-sided limits diverge to $\infty$ or $-\infty$
   • Example: $1/x$ at $x=0$

Standard Continuous Functions:

• Polynomials: continuous everywhere
• Rational functions: continuous wherever denominator $\neq 0$
• $\sin x, \cos x$: continuous everywhere
• $e^x, \ln x$: continuous on their domains

⚡ JEE Tip: To check continuity of $f(x) = \frac{g(x)}{h(x)}$ at $x=a$ where $h(a) = 0$, first check if $g(a) = 0$. If both are zero, factor and cancel, or use limit.

⚡ Common Mistake: A function can be discontinuous even if its derivative exists (e.g., $f(x) = |x|$ is continuous but not differentiable at $0$ — actually check: $|x|$ IS continuous at $0$). Example of continuous but not differentiable: $f(x) = |x|$ at $0$ IS differentiable from neither side… wait, $f(x) = |x|$ is continuous at 0, derivative doesn’t exist there. For a function that’s continuous but not differentiable, think $f(x) = x^{1/3}$ at $0$ or the famous Weierstrass function (nowhere differentiable but continuous everywhere).

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Continuity on an Interval:

• $f$ is continuous on $(a,b)$ if continuous at every point in $(a,b)$
• $f$ is continuous on $[a,b]$ if continuous on $(a,b)$ and $\lim_{x \to a^+} f(x) = f(a)$ and $\lim_{x \to b^-} f(x) = f(b)$

Intermediate Value Theorem (IVT): If $f$ is continuous on $[a,b]$ and $k$ is between $f(a)$ and $f(b)$, then there exists $c \in [a,b]$ such that $f(c) = k$.

Properties:

If $f$ and $g$ are continuous at $a$, then:

• $f+g$ is continuous at $a$
• $f \cdot g$ is continuous at $a$
• $\frac{f}{g}$ is continuous at $a$ if $g(a) \neq 0$

Composite Function: If $f$ is continuous at $a$ and $g$ is continuous at $f(a)$, then $g \circ f$ is continuous at $a$.

Uniform Continuity:

$f$ is uniformly continuous on $I$ if $\forall \epsilon > 0, \exists \delta > 0$ such that $|x-y| < delta$ implies $|f(x)-f(y)| < \epsilon$.

Key distinction: $\delta$ depends only on $\epsilon$ (not on $x$). Uniform continuity on a closed bounded interval implies continuity.

Worked Examples:

Example 1: Check continuity of $f(x) = \begin{cases} x^2 & x < 2 \ 3x-2 & x \geq 2 \end{cases}$ at $x=2$.

$\lim_{x \to 2^-} f(x) = 2^2 = 4$. $\lim_{x \to 2^+} f(x) = 3(2)-2 = 4$. $f(2) = 3(2)-2 = 4$. Since all three are 4, $f$ is continuous at $x=2$.

Example 2 (JEE 2021): For what value of $k$ is $f(x) = \begin{cases} \frac{\sin 3x}{x} & x \neq 0 \ k & x = 0 \end{cases}$ continuous at $x=0$?

We need $\lim_{x \to 0} \frac{\sin 3x}{x} = f(0) = k$. $\lim_{x \to 0} \frac{\sin 3x}{x} = 3 \cdot \frac{\sin 3x}{3x} \to 3 \cdot 1 = 3$. So $k = 3$.

Example 3: Show that $f(x) = \frac{1}{x}$ is not uniformly continuous on $(0,1)$.

Take $\epsilon = 1$. For any $\delta > 0$, choose $n$ large so $1/n < \delta$. Let $x = 1/(2n), y = 1/n$. Then $|x-y| = 1/(2n) < \delta$ for large $n$. $|f(x)-f(y)| = |2n - n| = n \geq 1$. Since $\delta$ can be made arbitrarily small, this violates uniform continuity.

Note: $f(x) = 1/x$ IS continuous on $(0,1)$ but NOT uniformly continuous.

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory for serious JEE Advanced preparation.

Continuity and Differentiability:

If $f$ is differentiable at $a$, then $f$ is continuous at $a$. Converse is false: $f(x) = |x|$ is continuous at $0$ but not differentiable at $0$.

Types of Discontinuities in Classification:

1. Limitable discontinuities (both one-sided limits exist):

   • Removable (limit = finite value)
   • Jump (finite but unequal one-sided limits)

2. Non-limitable discontinuities (at least one one-sided limit is infinite):

   • Infinite discontinuity
   • Oscillatory discontinuity (e.g., $\sin(1/x)$ at $0$)

Special Limits and Continuity:

$f(x) = \sin(1/x)$ for $x \neq 0$ and $f(0) = 0$:

• One-sided limits do not exist (oscillatory)
• Not removable
• The limit $\lim_{x \to 0} \sin(1/x)$ does not exist (oscillates between -1 and 1)

Weierstrass Function: $f(x) = \sum_{n=0}^{\infty} a^n \cos(b^n \pi x)$ with $0 < a < 1$ and $b$ odd integer $> 1/a$. This function is continuous everywhere but differentiable nowhere.

Advanced Problems:

Problem (JEE Advanced 2023): Let $f: \mathbb{R} \to \mathbb{R}$ be continuous and $f(f(x)) = x$ for all $x \in \mathbb{R}$. Prove that there exists $c \in \mathbb{R}$ such that $f(c) = c$.

We know $f$ is continuous and involutive ($f = f^{-1}$). Consider $g(x) = f(x) - x$. If $g(x) > 0$ for all $x$, then $f(x) > x$ for all $x$. But then $f(f(x)) > f(x) > x$, contradicting $f(f(x)) = x$. Similarly $g(x) < 0$ for all $x$ gives contradiction. So by IVT (since $g$ is continuous), there exists $c$ with $g(c) = 0$, i.e., $f(c) = c$.

Problem: Prove that every polynomial of odd degree has at least one real root.

Let $P(x)$ be polynomial of odd degree $n$. As $x \to \infty$, $P(x) \sim a_n x^n$ where $a_n > 0$ (or $< 0$ depending). So $\lim_{x \to \infty} P(x) = \infty$ and $\lim_{x \to -\infty} P(x) = -\infty$ (for $a_n > 0$). Since $P$ is continuous (polynomial), by IVT there exists $c$ with $P(c) = 0$.

JEE Advanced Patterns (2018–2024):

• IVT applications are very common in proving existence of roots
• Continuity of composite functions tested in 2020
• Continuity with parameters (finding $k$ for continuity) is frequent
• Uniform continuity questions are rare but appear in tough sets
• Discontinuity classification appeared in 2019, 2022

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