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Limits, Continuity and Differentiability: Mathematics | JEE Main

Q: Can we interchange the limit and the greatest integer function: is ?

Not in general, because is discontinuous at integers! However, , which is strictly between 0 and 1 (not an integer). In an open neighborhood around , takes values in , so identically. Thus the limit is 0.

Let

x=2

be a root of the equation

x^2 + px + q = 0

and

f(x) = \begin{cases} \frac{1-\cos(x^2 - 4px + q^2 + 8q + 16)}{(x-2p)^4} &, x \neq 2p \\ 0 &, x = 2p \end{cases}

\lim_{x\rightarrow 2p^+} [f(x)]

where

[\cdot]

denotes greatest integer function, is

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Step-by-step solutionView

Correct answer

Using $x=2$ as a root gives $q = -(2p+4)$ , simplifying the argument inside the cosine to $(x-2p)^2$ , which yields $\lim_{x\to 2p^+} f(x) = \frac{1}{2}$ and hence $[f(x)] = 0$ .

Option analysis

Why each option works or fails

A ·

0

None. The student correctly evaluates $\lim_{x\to 2p^+} f(x) = \frac{1}{2}$ and takes the floor to get $0$ . Correctly substitute $q = -2p - 4$ to rewrite the argument inside the cosine as $(x-2p)^2$ , apply the standard limit $\lim_{\theta\to 0}\frac{1-\cos\theta}{\theta^2} = \frac{1}{2}$ , and evaluate $[1/2] = 0$ as $f(x)$ approaches $1/2$ strictly from below or above within $(0, 1)$ .

B ·

-1

The student mistakenly believes that $f(x)$ approaches a negative value near $0$ , causing the floor function to yield $-1$ . Note that $1 - \cos(u) \ge 0$ for all real $u$ , and $(x-2p)^4 > 0$ , so $f(x) > 0$ everywhere near $x=2p$ . Thus $[f(x)]$ cannot be negative.

C ·

2

The student inverted the standard limit, using $\frac{1-\cos\theta}{\theta^2} \to 2$ instead of $\frac{1}{2}$ . Recall the standard Taylor expansion or half-angle identity: $1-\cos\theta = 2\sin^2(\theta/2) \approx \frac{\theta^2}{2}$ , so the limit is $\frac{1}{2}$ , not $2$ .

D ·

1

The student rounds the limit $\frac{1}{2}$ up to $1$ , or evaluates the limit after mistakenly concluding the value equals $1$ . The greatest integer function $[\cdot]$ rounds down to the nearest integer less than or equal to the input; for values strictly between $0$ and $1$ , $[f(x)] = 0$ .

Reviewed route

Solution

StepWorking

01given

$x=2$ is a root of $x^2 + px + q = 0$ , and $f(x) = \frac{1-\cos(x^2 - 4px + q^2 + 8q + 16)}{(x-2p)^4}$ for $x \neq 2p$ .

02goal

Evaluate the limit $\lim_{x \rightarrow 2p^+} [f(x)]$ , where $[\cdot]$ is the greatest integer function.

03approach

Use the root condition to express $q^2 + 8q + 16$ in terms of $p$ . Then, recognize the argument of the cosine as a perfect square $(x-2p)^2$ , apply the standard limit $\lim_{\theta \to 0} \frac{1-\cos\theta}{\theta^2} = \frac{1}{2}$ , and determine the floor of the approaching values.

04execute

Since $x=2$ is a root: $2^2 + 2p + q = 0 \implies 4 + 2p + q = 0 \implies q + 4 = -2p$ . Squaring both sides yields $(q + 4)^2 = q^2 + 8q + 16 = 4p^2$ .

05execute

Substitute $q^2 + 8q + 16 = 4p^2$ into $f(x)$ : $x^2 - 4px + q^2 + 8q + 16 = x^2 - 4px + 4p^2 = (x - 2p)^2$ . Therefore, $f(x) = \frac{1 - \cos((x-2p)^2)}{\big((x-2p)^2\big)^2}$ .

06execute

Let $\theta = (x-2p)^2$ . As $x \to 2p^+$ , $\theta \to 0^+$ . Using the standard limit, $\lim_{\theta \to 0^+} \frac{1 - \cos\theta}{\theta^2} = \frac{1}{2}$ . For small $\theta > 0$ , since $\cos\theta = 1 - \frac{\theta^2}{2} + \frac{\theta^4}{24} - \dots$ , we have $f(x) = \frac{1}{2} - \frac{\theta^2}{24} + \dots < \frac{1}{2}$ , and clearly $f(x) > 0$ . Thus, $0 < f(x) < 1$ , which gives $[f(x)] = 0$ . Hence $\lim_{x \to 2p^+} [f(x)] = [0] = 0$ .

✓verify

Check with a simple test value: choose $p = -2$ , then $q = -4 - 2(-2) = 0$ . Root condition: $2^2 - 2(2) + 0 = 0$ , verified. Then $2p = -4$ . $f(x) = \frac{1-\cos(x^2 + 8x + 16)}{(x+4)^4} = \frac{1-\cos(x+4)^2}{(x+4)^4} \to 1/2$ . For all $x$ near $-4$ , $f(x) \in (0, 1/2)$ , so $[f(x)] = 0$ .

Hints that build this answer step by step

Since $x=2$ is a root of $x^2 + px + q = 0$ , how can $q$ be expressed in terms of $p$ ?
$q = -2p - 4$
Substitute $q = -(2p+4)$ into the expression inside the cosine: $E = x^2 - 4px + q^2 + 8q + 16$ . What does it simplify to?
$(x - 2p)^2$
With $E = (x-2p)^2$ , what is $\lim_{x\to 2p^+} f(x) = \lim_{x\to 2p^+} \frac{1-\cos((x-2p)^2)}{(x-2p)^4}$ ?
$\frac{1}{2}$
Since $f(x) \to \frac{1}{2}$ as $x \to 2p^+$ , what is the value of $\lim_{x\to 2p^+} [f(x)]$ ?
$0$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Can we interchange the limit and the greatest integer function: is

\lim [f(x)] = [\lim f(x)]

Not in general, because $[\cdot]$ is discontinuous at integers! However, $\lim f(x) = 1/2$ , which is strictly between 0 and 1 (not an integer). In an open neighborhood around $2p$ , $f(x)$ takes values in $(0, 1)$ , so $[f(x)] = 0$ identically. Thus the limit is 0.

Answer

Using $x=2$ as a root gives $q = -(2p+4)$ , simplifying the argument inside the cosine to $(x-2p)^2$ , which yields $\lim_{x\to 2p^+} f(x) = \frac{1}{2}$ and hence $[f(x)] = 0$ .

Why each option works or fails

A · correct: $0$ - None. The student correctly evaluates $\lim_{x\to 2p^+} f(x) = \frac{1}{2}$ and takes the floor to get $0$ . Correctly substitute $q = -2p - 4$ to rewrite the argument inside the cosine as $(x-2p)^2$ , apply the standard limit $\lim_{\theta\to 0}\frac{1-\cos\theta}{\theta^2} = \frac{1}{2}$ , and evaluate $[1/2] = 0$ as $f(x)$ approaches $1/2$ strictly from below or above within $(0, 1)$ .
B: $-1$ - The student mistakenly believes that $f(x)$ approaches a negative value near $0$ , causing the floor function to yield $-1$ . Note that $1 - \cos(u) \ge 0$ for all real $u$ , and $(x-2p)^4 > 0$ , so $f(x) > 0$ everywhere near $x=2p$ . Thus $[f(x)]$ cannot be negative.
C: $2$ - The student inverted the standard limit, using $\frac{1-\cos\theta}{\theta^2} \to 2$ instead of $\frac{1}{2}$ . Recall the standard Taylor expansion or half-angle identity: $1-\cos\theta = 2\sin^2(\theta/2) \approx \frac{\theta^2}{2}$ , so the limit is $\frac{1}{2}$ , not $2$ .
D: $1$ - The student rounds the limit $\frac{1}{2}$ up to $1$ , or evaluates the limit after mistakenly concluding the value equals $1$ . The greatest integer function $[\cdot]$ rounds down to the nearest integer less than or equal to the input; for values strictly between $0$ and $1$ , $[f(x)] = 0$ .

Step-by-step solution

given: $x=2$ is a root of $x^2 + px + q = 0$ , and $f(x) = \frac{1-\cos(x^2 - 4px + q^2 + 8q + 16)}{(x-2p)^4}$ for $x \neq 2p$ .
goal: Evaluate the limit $\lim_{x \rightarrow 2p^+} [f(x)]$ , where $[\cdot]$ is the greatest integer function.
approach: Use the root condition to express $q^2 + 8q + 16$ in terms of $p$ . Then, recognize the argument of the cosine as a perfect square $(x-2p)^2$ , apply the standard limit $\lim_{\theta \to 0} \frac{1-\cos\theta}{\theta^2} = \frac{1}{2}$ , and determine the floor of the approaching values.
execute: Since $x=2$ is a root: $2^2 + 2p + q = 0 \implies 4 + 2p + q = 0 \implies q + 4 = -2p$ . Squaring both sides yields $(q + 4)^2 = q^2 + 8q + 16 = 4p^2$ .
execute: Substitute $q^2 + 8q + 16 = 4p^2$ into $f(x)$ : $x^2 - 4px + q^2 + 8q + 16 = x^2 - 4px + 4p^2 = (x - 2p)^2$ . Therefore, $\lim_{x \rightarrow 2p^+} [f(x)]$ .
execute: Let $\theta = (x-2p)^2$ . As $x \to 2p^+$ , $\theta \to 0^+$ . Using the standard limit, $\lim_{\theta \to 0^+} \frac{1 - \cos\theta}{\theta^2} = \frac{1}{2}$ . For small $\theta > 0$ , since $\cos\theta = 1 - \frac{\theta^2}{2} + \frac{\theta^4}{24} - \dots$ , we have $f(x) = \frac{1}{2} - \frac{\theta^2}{24} + \dots < \frac{1}{2}$ , and clearly $f(x) > 0$ . Thus, $0 < f(x) < 1$ , which gives $[f(x)] = 0$ . Hence $\lim_{x \to 2p^+} [f(x)] = [0] = 0$ .
verify: Check with a simple test value: choose $p = -2$ , then $q = -4 - 2(-2) = 0$ . Root condition: $2^2 - 2(2) + 0 = 0$ , verified. Then $2p = -4$ . $f(x) = \frac{1-\cos(x^2 + 8x + 16)}{(x+4)^4} = \frac{1-\cos(x+4)^2}{(x+4)^4} \to 1/2$ . For all $x$ near $-4$ , $f(x) \in (0, 1/2)$ , so $[f(x)] = 0$ .

Limits, Continuity and Differentiability: Mathematics | JEE Main

What feels right?

Using $x=2$ as a root gives $q = -(2p+4)$ , simplifying the argument inside the cosine to $(x-2p)^2$ , which yields $\lim_{x\to 2p^+} f(x) = \frac{1}{2}$ and hence $[f(x)] = 0$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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