Greatest Integer Function in Limits & Continuity

Introduction

The Greatest Integer Function (GIF), denoted $\lfloor x \rfloor$ or $[x]$, returns the largest integer less than or equal to $x$. JEE Main tests GIF-based limits and continuity problems in 20–25 questions over the past six years. These problems are uniquely tricky because GIF creates discontinuities at every integer, and left-hand and right-hand limits behave very differently. This article gives you the complete strategy.

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1. Fundamental Properties

Definition

$\lfloor x \rfloor = \text{greatest integer } n \text{ such that } n \leq x.$

Examples: $\lfloor 3.7 \rfloor = 3$, $\lfloor -1.3 \rfloor = -2$, $\lfloor 5 \rfloor = 5$.

Essential Properties

Behaviour Near Integers

For any integer $n$: $\lim_{x \to n^-} \lfloor x \rfloor = n - 1, \qquad \lim_{x \to n^+} \lfloor x \rfloor = n.$

So $\lfloor x \rfloor$ is right-continuous but left-discontinuous at every integer.

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2. Limits Involving GIF

The Squeeze (Sandwich) Theorem Approach

Since $x - 1 < \lfloor x \rfloor \leq x$, we can squeeze GIF-containing expressions: $\frac{x - 1}{x} < \frac{\lfloor x \rfloor}{x} \leq 1 \quad \text{for } x > 0.$ As $x \to \infty$: $\lim_{x \to \infty} \frac{\lfloor x \rfloor}{x} = 1.$

Standard GIF Limit Results

$\lim_{x \to 0^+} x \lfloor 1/x \rfloor = 1, \qquad \lim_{x \to 0^+} \frac{\lfloor x \rfloor}{x} = 0, \qquad \lim_{x \to 0^-} \frac{\lfloor x \rfloor}{x} = \text{undefined (diverges)}.$

Warning: $\lim \lfloor f(x) \rfloor \neq \lfloor \lim f(x) \rfloor$ in General

Example: $\lim_{x \to 0^-} \lfloor x \rfloor = -1$, but $\lfloor \lim_{x \to 0^-} x \rfloor = \lfloor 0 \rfloor = 0$.

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3. Worked Examples

Example 1 (Basic Continuity Check)

Discuss the continuity of $f(x) = \lfloor x \rfloor + \lfloor -x \rfloor$.

Solution:

Case 1: $x \in \mathbb{Z}$. Then $f(x) = x + (-x) = 0$.

Case 2: $x \notin \mathbb{Z}$. Let $x = n + \alpha$ where $0 < \alpha < 1$. Then $\lfloor x \rfloor = n$ and $\lfloor -x \rfloor = -(n+1)$. So $f(x) = n - n - 1 = -1$.

Thus: $f(x) = \begin{cases} 0, & x \in \mathbb{Z} \\ -1, & x \notin \mathbb{Z} \end{cases}$

At any integer $n$: $\lim_{x \to n} f(x) = -1 \neq 0 = f(n)$.

$f$ is discontinuous at every integer.

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Example 2 (Limit with GIF in Exponent)

Evaluate $\displaystyle \lim_{x \to 0^+} (1+x)^{\lfloor 1/x \rfloor}$.

Solution:

As $x \to 0^+$, $\lfloor 1/x \rfloor$ behaves like $1/x$ (since $\lfloor 1/x \rfloor \approx 1/x - \{1/x\}$ and $\{1/x\}$ is bounded).

More precisely, $\frac{1}{x} - 1 < \lfloor 1/x \rfloor \leq \frac{1}{x}$.

$(1+x)^{1/x - 1} < (1+x)^{\lfloor 1/x \rfloor} \leq (1+x)^{1/x}.$

Both bounds tend to $e$:

• $(1+x)^{1/x} \to e$.
• $(1+x)^{1/x - 1} = \frac{(1+x)^{1/x}}{1+x} \to \frac{e}{1} = e$.

By squeeze: the limit is $e$.

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Example 3 (Continuity of $x\lfloor x \rfloor$)

Is $f(x) = x \lfloor x \rfloor$ continuous at $x = 2$?

Solution:

$f(2) = 2 \cdot 2 = 4$.

LHL: $\lim_{x \to 2^-} x \lfloor x \rfloor = 2 \cdot 1 = 2$ (since $\lfloor x \rfloor = 1$ for $1 \leq x < 2$).

RHL: $\lim_{x \to 2^+} x \lfloor x \rfloor = 2 \cdot 2 = 4$.

LHL $\neq$ RHL. Not continuous at $x = 2$.

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Example 4 (Fractional Part Limit)

Evaluate $\displaystyle \lim_{x \to 3^-} \frac{\{x\}}{x - \lfloor x \rfloor}$.

Solution:

Since $\{x\} = x - \lfloor x \rfloor$, this is: $\lim_{x \to 3^-} \frac{x - \lfloor x \rfloor}{x - \lfloor x \rfloor} = 1.$

But what about $\lim_{x \to 3^+}$? Same thing: $\{x\}/(x - \lfloor x \rfloor) = 1$ for all non-integer $x$.

The function equals 1 everywhere except at integers where it is $0/0$ (undefined).

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Example 5 (GIF Inside Trigonometric Function)

Find $\displaystyle \lim_{x \to 0} \frac{\sin \lfloor x \rfloor}{\lfloor x \rfloor}$.

Solution:

RHL ($x \to 0^+$): $\lfloor x \rfloor = 0$, so $\frac{\sin 0}{0} = \frac{0}{0}$, which is indeterminate. But since $\lfloor x \rfloor = 0$ identically for $0 < x < 1$, the expression is $\frac{0}{0}$, which is undefined.

LHL ($x \to 0^-$): $\lfloor x \rfloor = -1$, so $\frac{\sin(-1)}{-1} = \frac{-\sin 1}{-1} = \sin 1$.

The limit does not exist (LHL exists but RHL is undefined).

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Example 6 (Summation with GIF)

Evaluate $\displaystyle \lim_{n \to \infty} \frac{1}{n^2} \sum_{r=1}^{n} r \lfloor r/n \rfloor$.

Solution:

For $1 \leq r \leq n-1$: $0 < r/n < 1$, so $\lfloor r/n \rfloor = 0$.

For $r = n$: $\lfloor n/n \rfloor = 1$.

So the sum $= 0 + 0 + \cdots + n \cdot 1 = n$. $\lim_{n \to \infty} \frac{n}{n^2} = 0.$

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4. Previous Year JEE Problems

Problem 1 (JEE Main 2023)

$\displaystyle \lim_{x \to 0} \frac{x^2}{\sin x \lfloor \tan x \rfloor}$ where $x \to 0^-$:

Solution:

As $x \to 0^-$, $\tan x \to 0^-$, so $\lfloor \tan x \rfloor = -1$. $\lim_{x \to 0^-} \frac{x^2}{\sin x \cdot (-1)} = \lim_{x \to 0^-} \frac{-x^2}{\sin x} = \lim_{x \to 0^-} \frac{-x}{\sin x / x} \cdot \frac{x}{1} = 0.$

Wait—more carefully: $\frac{x^2}{\sin x \cdot (-1)} = -\frac{x^2}{\sin x} = -x \cdot \frac{x}{\sin x} \to 0 \cdot 1 = 0.$

Answer: 0.

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Problem 2 (JEE Main 2021)

The number of points of discontinuity of $f(x) = \lfloor x^2 - x \rfloor$ in $[0, 2]$ is:

Solution:

$g(x) = x^2 - x$ on $[0, 2]$ has range: $g(0) = 0$, $g(1/2) = -1/4$, $g(1) = 0$, $g(2) = 2$.

The integers in the range $[-1/4, 2]$ are: $0, 1, 2$.

$g(x) = 0$: $x = 0, 1$. $g(x) = 1$: $x^2 - x - 1 = 0 \Rightarrow x = \frac{1+\sqrt{5}}{2} \approx 1.618$. $g(x) = 2$: $x = 2$.

Points of discontinuity: $x = 0$ (check left—not in domain), $x = 1$, $x = \frac{1+\sqrt{5}}{2}$, $x = 2$.

But $x = 0$ is the left endpoint, and $g$ passes through 0 at $x = 0, 1$. We need to check where $g(x)$ crosses an integer:

Between $x = 0$ and $x = 1$, $g(x)$ dips below 0 (minimum $-1/4$) so $\lfloor g(x) \rfloor$ jumps between 0 and $-1$. Discontinuity at points where $g(x) = 0$ in this interval — that's at $x = 0$ and $x = 1$.

After $x = 1$: $g(x)$ increases from 0 to 2. Discontinuity at $g(x) = 1$ and $g(x) = 2$.

Answer: 4 points of discontinuity.

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Problem 3 (JEE Main 2024)

If $f(x) = \lfloor 2\sin x \rfloor$ on $[0, \pi]$, at how many points is $f$ discontinuous?

Solution:

On $[0, \pi]$, $\sin x$ goes from 0 to 1 and back to 0. So $2\sin x$ ranges from 0 to 2 and back.

Values where $2\sin x$ equals an integer:

• $2\sin x = 0$: $x = 0, \pi$
• $2\sin x = 1$: $x = \pi/6, 5\pi/6$
• $2\sin x = 2$: $x = \pi/2$

The function $\lfloor 2\sin x \rfloor$ changes value at each of these (except at $x = \pi/2$ where $2\sin x$ touches 2 from below).

At $x = \pi/2$: $\lfloor 2\sin x \rfloor = 1$ just before and after (since $2\sin x < 2$ nearby), and $\lfloor 2 \rfloor = 2$ at $x = \pi/2$. So this is a discontinuity.

Points of discontinuity: $x = \pi/6$ (jumps from 0 to 1), $x = \pi/2$ (isolated point at 2), $x = 5\pi/6$ (drops from 1 to 0).

Answer: 3.

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5. Common Patterns & Quick Results

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6. Tips for JEE Aspirants

1. Always compute LHL and RHL separately at integer points—GIF is discontinuous at every integer.
2. For $\lfloor f(x) \rfloor$, find where $f(x)$ takes integer values—these are the candidate discontinuity points.
3. Use the squeeze theorem for limits at infinity: $x - 1 < \lfloor x \rfloor \leq x$ is your best friend.
4. Fractional part $\{x\} = x - \lfloor x \rfloor$ simplifies many expressions. At non-integer points, this is straightforward.
5. Draw the staircase graph mentally. Visualizing the step function makes the problem transparent.
6. For composite GIF $\lfloor g(x) \rfloor$, solve $g(x) = n$ for each relevant integer $n$ to find all discontinuity points.

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7. Quick-Reference Summary

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GIF problems reward students who think carefully about one-sided limits. The function is simple, but the traps are everywhere—practice until the staircase behaviour becomes second nature.