Continuity of Piecewise Functions: Parameter Determination

Introduction

One of the most heavily tested topics in JEE Main is determining unknown parameters so that a piecewise-defined function becomes continuous (or continuous and differentiable). With 55–60 questions over the past six years, this is the single most asked sub-topic within Limits, Continuity, and Differentiability. The key idea is deceptively simple—equate left-hand and right-hand limits to the function value at the junction point—but the algebra can get tricky when multiple parameters, trigonometric identities, or indeterminate forms are involved.

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1. Fundamental Principle

Definition of Continuity at a Point

A function $f$ is continuous at $x = a$ if and only if all three conditions hold:

1. $f(a)$ is defined.
2. $\displaystyle \lim_{x \to a} f(x)$ exists, i.e., $\displaystyle \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$.
3. $\displaystyle \lim_{x \to a} f(x) = f(a)$.

For Piecewise Functions

If $f(x) = \begin{cases} g(x), & x < a \\ k, & x = a \\ h(x), & x > a \end{cases}$ then continuity at $x = a$ requires: $\boxed{\lim_{x \to a^-} g(x) = k = \lim_{x \to a^+} h(x)}$

Three-Piece Pattern

When the function has three or more pieces with two junction points $a$ and $b$, apply the continuity condition independently at each junction. This typically gives a system of equations in the unknown parameters.

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2. Step-by-Step Method

Step 1: Identify the junction points—the values of $x$ where the rule changes.

Step 2: At each junction point $x = a$, compute:

• Left-hand limit: $\displaystyle \lim_{x \to a^-} f(x)$
• Right-hand limit: $\displaystyle \lim_{x \to a^+} f(x)$
• Function value: $f(a)$

Step 3: Set LHL = RHL = $f(a)$. If the limits produce $\frac{0}{0}$ or other indeterminate forms, resolve them using L'Hôpital's rule, standard limits, or algebraic manipulation before equating.

Step 4: Solve the resulting system of equations for the unknown parameters.

Step 5: Verify—substitute back and confirm the limits match.

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3. Worked Examples

Example 1 (Single Junction, Trigonometric)

Let f(x) = \begin{cases} \dfrac{\sin(a+1)x + \sin x}{x}, & x < 0 \6pt] c, & x = 0 $6pt] \dfrac{\sqrt{x+bx^2} - \sqrt{x}}{bx^{3/2}}, & x > 0 \end{cases} $Find$ a, b, c $so that$ f $is continuous at$ x = 0 $.

Solution:

LHL ($x \to 0^-$): $\lim_{x \to 0^-} \frac{\sin(a+1)x + \sin x}{x} = (a+1) + 1 = a + 2.$ (Using $\lim_{x \to 0} \frac{\sin kx}{x} = k$.)

RHL ($x \to 0^+$): $\lim_{x \to 0^+} \frac{\sqrt{x+bx^2} - \sqrt{x}}{bx^{3/2}} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{1+bx} - 1)}{bx^{3/2}} = \lim_{x \to 0^+} \frac{\sqrt{1+bx} - 1}{bx}.$ Using $\sqrt{1+u} \approx 1 + \frac{u}{2}$ for small $u$: $= \lim_{x \to 0^+} \frac{\frac{bx}{2}}{bx} = \frac{1}{2}.$

Setting LHL = $c$ = RHL: $a + 2 = c = \frac{1}{2} \quad \Rightarrow \quad a = -\frac{3}{2}, \quad c = \frac{1}{2}.$ $b$ can be any non-zero real number (it cancels out).

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Example 2 (Two Parameters, Two Junctions)

Let f(x) = \begin{cases} \dfrac{x^2 - (a+2)x + a}{x - 2}, & x \neq 2 \6pt] 2, & x = 2 \end{cases} $be continuous at$ x = 2 $. Find$ a $.

Solution:

At $x = 2$, the denominator is zero, so we need the numerator to also be zero for the limit to exist: $4 - 2(a+2) + a = 0 \quad \Rightarrow \quad 4 - 2a - 4 + a = 0 \quad \Rightarrow \quad a = 0.$ So $f(x) = \frac{x^2 - 2x}{x - 2} = \frac{x(x-2)}{x-2} = x$ for $x \neq 2$.

Check: $\lim_{x \to 2} f(x) = 2 = f(2)$. Continuous.

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Example 3 (Exponential $1^\infty$ Form)

Let f(x) = \begin{cases} (1 + |\sin x|)^{\frac{a}{|\sin x|}}, & -\frac{\pi}{6} < x < 0 \6pt] b, & x = 0 $6pt] e^{\frac{\tan 2x}{\tan 3x}}, & 0 < x < \frac{\pi}{6} \end{cases} $Determine$ a, b $for continuity at$ x = 0 $.

Solution:

LHL ($x \to 0^-$): Let $t = |\sin x|$. As $x \to 0$, $t \to 0^+$. $\lim_{t \to 0^+} (1+t)^{a/t} = \left[ \lim_{t \to 0^+} (1+t)^{1/t} \right]^a = e^a.$

RHL ($x \to 0^+$): $\lim_{x \to 0^+} e^{\frac{\tan 2x}{\tan 3x}} = \exp\left( \lim_{x \to 0^+} \frac{\tan 2x}{2x} \cdot \frac{3x}{\tan 3x} \cdot \frac{2}{3} \right) = e^{2/3}.$

For continuity, LHL = $f(0)$ = RHL: $e^a = b = e^{2/3}.$ Thus, $a = \frac{2}{3}$ and $b = e^{2/3}$.

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Example 4 (Absolute Value at Junction)

Let f(x) = \begin{cases} a + bx, & x < 1 \6pt] 4, & x = 1 $6pt] b - ax, & x > 1 \end{cases} $For$ f $to be continuous at$ x = 1 $, find$ a $and$ b $.

Solution:

LHL: $a + b(1) = a + b$. RHL: $b - a(1) = b - a$. $f(1) = 4$.

Set LHL = RHL = 4: $a + b = 4 \quad \text{and} \quad b - a = 4.$ Adding: $2b = 8 \Rightarrow b = 4$. Subtracting: $2a = 0 \Rightarrow a = 0$.

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Example 5 (Trigonometric with 0/0)

If $f(x)$ is continuous at $x = 0$, where f(x) = \begin{cases} \dfrac{1 - \cos 4x}{x^2}, & x < 0 \6pt] a, & x = 0 $6pt] \dfrac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & x > 0 \end{cases} $find$ a $.

Solution:

LHL: $\lim_{x \to 0^-} \frac{1 - \cos 4x}{x^2} = \lim_{x \to 0} \frac{2\sin^2 2x}{x^2} = 2 \cdot \lim_{x \to 0} \frac{\sin^2 2x}{x^2} = 2 \cdot 4 = 8.$

RHL: Rationalize by multiplying numerator and denominator by $\sqrt{16+\sqrt{x}} + 4$: $\lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}} + 4)}{16+\sqrt{x} - 16} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}} + 4)}{\sqrt{x}} = \sqrt{16} + 4 = 8.$

Thus $a = 8$.

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Example 6 (Exponential 0/0 Form)

Let f(x) = \begin{cases} \dfrac{e^{3x} - 1}{\log(1 + 5x)}, & x \neq 0 \6pt] k, & x = 0 \end{cases} $Find$ k $so that$ f $is continuous at$ x = 0 $.

Solution:

Both numerator and denominator tend to 0 as $x \to 0$. Using standard limits: $\lim_{x \to 0} \frac{e^{3x} - 1}{\log(1+5x)} = \lim_{x \to 0} \frac{\frac{e^{3x}-1}{x}}{\frac{\log(1+5x)}{x}} = \frac{3}{5}.$

So $k = \frac{3}{5}$.

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4. Previous Year JEE Problems

Problem 1 (JEE Main 2023)

If the function f(x) = \begin{cases} \dfrac{\sin(a+2)x + \sin x}{x}, & x < 0 \6pt] b, & x = 0 $6pt] \dfrac{(x + 3x^2)^{1/3} - x^{1/3}}{x^{4/3}}, & x > 0 \end{cases} $is continuous at$ x = 0 $, then$ a + 2b $ is equal to:

Solution:

LHL: $\lim_{x \to 0^-} \frac{\sin(a+2)x + \sin x}{x} = (a+2) + 1 = a + 3$.

RHL: Factor $x^{1/3}$ from the numerator: $\lim_{x \to 0^+} \frac{x^{1/3}[(1+3x)^{1/3} - 1]}{x^{4/3}} = \lim_{x \to 0^+} \frac{(1+3x)^{1/3} - 1}{x}.$ Using $(1+u)^{1/3} \approx 1 + \frac{u}{3}$: the limit $= \frac{3}{3} = 1$.

So $a + 3 = b = 1$, giving $a = -2, b = 1$.

$a + 2b = -2 + 2 = 0$.

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Problem 2 (JEE Main 2020)

If the function f(x) = \begin{cases} \dfrac{\sqrt{1+kx} - \sqrt{1-kx}}{x}, & -1 \le x < 0 \6pt] 2x^2 + 3x - 2, & 0 \le x \le 1 \end{cases} $is continuous at$ x = 0 $, find$ k $.

Solution:

LHL ($x \to 0^-$): Rationalize the numerator: $\lim_{x \to 0^-} \frac{(1+kx) - (1-kx)}{x(\sqrt{1+kx} + \sqrt{1-kx})} = \lim_{x \to 0^-} \frac{2kx}{x(\sqrt{1+kx} + \sqrt{1-kx})}$ $= \frac{2k}{\sqrt{1} + \sqrt{1}} = \frac{2k}{2} = k.$

RHL ($x \to 0^+$) and $f(0)$: $\lim_{x \to 0^+} (2x^2 + 3x - 2) = -2.$

Since $f$ is continuous at $x = 0$, LHL = RHL: $k = -2.$

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Problem 3 (JEE Main 2024)

If $f(x)$ is continuous at $x = 0$, where f(x) = \begin{cases} \dfrac{e^{2x} - 1 - 2x}{x^2}, & x \neq 0 \6pt] k, & x = 0 \end{cases} $find$ k $.

Solution:

Apply L'Hôpital's rule (since both numerator and denominator $\to 0$): $\lim_{x \to 0} \frac{e^{2x} - 1 - 2x}{x^2} = \lim_{x \to 0} \frac{2e^{2x} - 2}{2x} = \lim_{x \to 0} \frac{4e^{2x}}{2} = 2.$

Alternatively, using Taylor expansion: $e^{2x} = 1 + 2x + \frac{4x^2}{2} + \cdots$, so $\frac{e^{2x} - 1 - 2x}{x^2} = \frac{2x^2 + \cdots}{x^2} \to 2.$

Hence $k = 2$.

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5. Common Patterns & Quick Results

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6. Tips for JEE Aspirants

1. Always check both sides independently. Never assume LHL = RHL without computing both—especially with modulus, GIF, or exponential functions.
2. Use standard limit results to quickly evaluate $\frac{0}{0}$ forms rather than L'Hôpital every time—it is faster.
3. Factor before canceling. If the denominator has $(x - a)$, ensure the numerator also has this factor, then cancel.
4. System of equations. With two junctions and two unknowns, you get a solvable $2 \times 2$ system—solve efficiently by adding/subtracting the equations.
5. Watch for "automatic" continuity. Polynomials, $e^x$, $\sin x$, $\cos x$ are continuous everywhere—the issue arises only at junction points.
6. Verify your answer. Plug back and check both limits match $f(a)$. A 10-second verification can save you from sign errors.

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7. Quick-Reference Summary

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Piecewise continuity problems are the bread and butter of JEE Main Limits questions. Master the standard limit results in the table above, and you can solve most of these in under 2 minutes.