JEE Advanced 1985

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f (x) = \frac{s i n [ x ]}{[ x ]}, [x] \neq = 0$ ; $f (x) = 0, [x] = 0$ . Where $[x]$ denotes the greatest integer less than or equal to $x$ . then $lim_{x \to 0} f (x)$ equals

Q: Why can't we just use the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ here?

Great observation, but be careful! The standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ applies to the variable $x$ itself. Here, we have the Greatest Integer Function $[x]$ inside the sine and the denominator. The function is constant over intervals, not continuous, so the standard limit theorem doesn't apply directly.

Q: Is it possible for the limit to exist if the function is defined as $0$ at $x=0$?

The value of the function at the point $f(0)$ is irrelevant to the existence of the limit. The limit only cares about the behavior of the function *near* the point, not *at* the point. Even if $f(0)$ were defined differently, the LHL and RHL would still be unequal.

Select Answer:

Visualized Solution

The Piecewise Function

Given: $f (x) = \frac{s i n [ x ]}{[ x ]}$ for $[x] \neq = 0$
Given: $f (x) = 0$ for $[x] = 0$
Goal: Find $lim_{x \to 0} f (x)$

Condition for Limit Existence

For $lim_{x \to 0} f (x)$ to exist:
Left-Hand Limit (LHL) must equal Right-Hand Limit (RHL)
$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{+}} f (x)$

Setting up the Left-Hand Limit (LHL)

Let's evaluate the LHL: $lim_{x \to 0^{-}} f (x)$
Here, $x$ approaches $0$ from the left.
This means $x$ is slightly less than $0$ .

Evaluating $[x]$ for LHL

For $x \to 0^{-}$ , $x$ lies in the interval $(- 1, 0)$ .
By definition of GIF, the greatest integer $\leq x$ is $- 1$ .
Therefore, $[x] = - 1$ .

Substituting $[x]$ into $f (x)$

Since $[x] = - 1$ , we have $[x] \neq = 0$ .
We must use the first branch: $f (x) = \frac{s i n [ x ]}{[ x ]}$
Substitute $[x] = - 1$ : $f (x) = \frac{s i n ( - 1 )}{- 1}$

Simplifying the LHL

Recall the trigonometric identity: $sin (- θ) = - sin (θ)$
$f (x) = \frac{- s i n ( 1 )}{- 1}$
$LHL = sin (1)$

Setting up the Right-Hand Limit (RHL)

Now, let's evaluate the RHL: $lim_{x \to 0^{+}} f (x)$
Here, $x$ approaches $0$ from the right.
This means $x$ is slightly greater than $0$ .

Evaluating $[x]$ for RHL

For $x \to 0^{+}$ , $x$ lies in the interval $(0, 1)$ .
By definition of GIF, the greatest integer $\leq x$ is $0$ .
Therefore, $[x] = 0$ .

Evaluating $f (x)$ for RHL

Since $[x] = 0$ , we look at the second branch of the function.
The problem states: $f (x) = 0$ when $[x] = 0$ .
Therefore, $RHL = 0$ .

Comparing LHL and RHL

We found: $LHL = sin (1)$
We found: $RHL = 0$
Clearly, $sin (1) \neq = 0$ .
Therefore, $LHL \neq = RHL$ .

Final Conclusion

Since the Left-Hand Limit and Right-Hand Limit are not equal, the limit does not exist.
$lim_{x \to 0} f (x)$ does not exist.
Correct Option: (3) none of these

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The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Solution Diagram

Analyzing the Setup

We are tasked with evaluating the limit of the function

f (x)

x

approaches

0

, where the function is defined as:

f (x) = {\frac{s i n ([ x ])}{[ x ]} 0 if [x] e q 0 if [x] = 0

Here,

[x]

denotes the Greatest Integer Function. To determine if the limit exists, we must evaluate the Left-Hand Limit (LHL) and the Right-Hand Limit (RHL) independently.

The Left-Hand Approach

We approach

x = 0

from the left (

x \to 0^{-}

). In the interval

(- 1, 0)

, the value of the greatest integer function is constant:

[x] = - 1

Substituting this into our function definition, we observe:

f (x) = \frac{sin ( - 1 )}{- 1}

Using the odd property of the sine function,

sin (- θ) = - sin (θ)

, we simplify the expression:

f (x) = \frac{- sin ( 1 )}{- 1} = sin (1)

Thus, the Left-Hand Limit (LHL) is

sin (1)

The Right-Hand Approach

Next, we approach

x = 0

from the right (

x \to 0^{+}

). In the interval

(0, 1)

, the value of the greatest integer function is:

[x] = 0

According to the piecewise definition provided, when

[x] = 0

, the function value is explicitly defined as

0

. Therefore:

x \to 0^{+} lim f (x) = 0

Thus, the Right-Hand Limit (RHL) is

0

The Verdict

For a limit to exist at a point, the LHL and RHL must be equal. Comparing our results:

LHL = sin (1) \approx 0.841

RHL = 0

Since $\sin(1) eq 0$, the two paths do not converge to the same value. Consequently, the limit

lim_{x \to 0} f (x)

does not exist.

Common Traps & FAQs

Wait, why do we use

[x] = - 1

for the left-hand limit? Shouldn't it be

0

That is a classic trap! Remember, the Greatest Integer Function

[x]

returns the largest integer *less than or equal to*

x

. If

x

is slightly less than

0

(like

- 0.001

), the largest integer less than or equal to that value is indeed

- 1

, not

0

Why can't we just use the standard limit

lim_{x \to 0} \frac{s i n x}{x} = 1

here?

Great observation, but be careful! The standard limit

lim_{x \to 0} \frac{s i n x}{x} = 1

applies to the variable

x

itself. Here, we have the Greatest Integer Function

[x]

inside the sine and the denominator. The function is constant over intervals, not continuous, so the standard limit theorem doesn't apply directly.

If the RHL is

0

, why isn't the limit just

0

A limit only exists if the path from the left and the path from the right converge to the *same* value. Even if the RHL is

0

, if the LHL is

sin (1)

, the function is essentially 'breaking' at

x = 0

. Since the two sides don't meet, the limit cannot exist.

Is it possible for the limit to exist if the function is defined as

0

x = 0

The value of the function at the point

f (0)

is irrelevant to the existence of the limit. The limit only cares about the behavior of the function *near* the point, not *at* the point. Even if

f (0)

were defined differently, the LHL and RHL would still be unequal.

Why is the answer 'none of these' instead of 'does not exist'?

In many competitive exams, if the calculated result is 'does not exist' and that specific phrase isn't an option, 'none of these' is the correct choice. Always double-check the provided options against your derivation!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f (x) = \frac{s i n [ x ]}{[ x ]}, [x] \neq = 0$ ; $f (x) = 0, [x] = 0$ . Where $[x]$ denotes the greatest integer less than or equal to $x$ . then $lim_{x \to 0} f (x)$ equals

Select Answer:

Visualized Solution

The Piecewise Function

Condition for Limit Existence

Setting up the Left-Hand Limit (LHL)

Evaluating $[x]$ for LHL

Substituting $[x]$ into $f (x)$

Simplifying the LHL

Setting up the Right-Hand Limit (RHL)

Evaluating $[x]$ for RHL

Evaluating $f (x)$ for RHL

Comparing LHL and RHL

Final Conclusion

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Left-Hand Approach

The Right-Hand Approach

The Verdict

Similar Questions

For each $x \in R$ , let $[x]$ be the greatest integer less than or equal to $x$ . Then $lim_{x \to 0^{+}} \frac{x ([ x ] + ∣ x ∣ ) s i n [ x ]}{∣ x ∣}$ is equal to

If $f (x) = \frac{x - s i n x}{x + c o s ^{2} x}$ , then $lim_{x \to \infty} f (x)$ is

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ equals

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ is equal to

The value of $lim_{x \to 0} ((sin x)^{1/ x} + (1/ x)^{s i n x})$ , where $x > 0$ is

Let $[x]$ denote the greatest integer less than or equal to $x$ . Then $lim_{x \to 0} \frac{t a n ( π s i n ^{2} x ) + ( ∣ x ∣ - s i n ( x [ x ]) ) ^{2}}{x ^{2}}$ :

$lim_{x \to 0} \frac{e ^{4∣ s i n x ∣} - 2∣ s i n x ∣ - 1}{x ^{2}}$

For each $t \in R$ , let [t] be the greatest integer less than or equal to t. Then, $lim_{x \to 1^{+}} \frac{( 1 - ∣ x ∣ + s i n ∣1 - x ∣ ) s i n ( \frac{π}{2} [ 1 - x ])}{∣1 - x ∣ [ 1 - x ]}$

The value of $lim_{x \to 0^{+}} \frac{c o s ^{- 1} ( x - [ x ] ^{2} ) \cdot s i n ^{- 1} ( x - [ x ] ^{2} )}{x - x ^{3}}$ , where $[x]$ denotes the greatest integer $\leq x$ is :

Let $f : R \to R$ be a function defined as $f (x) = a sin (\frac{π [ x ]}{2}) + [2 - x], a \in R$ , where $[t]$ is the greatest integer less than or equal to $t$ . If $lim_{x \to - 1} f (x)$ exists, then the value of $\int_{0}^{4} f (x) d x$ is equal to :

Select Answer:

Visualized Solution

The Piecewise Function

Condition for Limit Existence

Setting up the Left-Hand Limit (LHL)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [x] for LHL

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [x] into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Simplifying the LHL

Setting up the Right-Hand Limit (RHL)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [x] for RHL

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) for RHL

Comparing LHL and RHL

Final Conclusion

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Left-Hand Approach

The Right-Hand Approach

The Verdict

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​x2sin(πcos2x)​ equals

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of limx→0+​x−x3cos−1(x−[x]2)⋅sin−1(x−[x]2)​, where [x] denotes the greatest integer ≤x is :

Evaluating $[x]$ for LHL

Substituting $[x]$ into $f (x)$

Evaluating $[x]$ for RHL

Evaluating $f (x)$ for RHL

For each $x \in R$ , let $[x]$ be the greatest integer less than or equal to $x$ . Then $lim_{x \to 0^{+}} \frac{x ([ x ] + ∣ x ∣ ) s i n [ x ]}{∣ x ∣}$ is equal to

If $f (x) = \frac{x - s i n x}{x + c o s ^{2} x}$ , then $lim_{x \to \infty} f (x)$ is

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ equals

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ is equal to

The value of $lim_{x \to 0} ((sin x)^{1/ x} + (1/ x)^{s i n x})$ , where $x > 0$ is

Let $[x]$ denote the greatest integer less than or equal to $x$ . Then $lim_{x \to 0} \frac{t a n ( π s i n ^{2} x ) + ( ∣ x ∣ - s i n ( x [ x ]) ) ^{2}}{x ^{2}}$ :

$lim_{x \to 0} \frac{e ^{4∣ s i n x ∣} - 2∣ s i n x ∣ - 1}{x ^{2}}$

For each $t \in R$ , let [t] be the greatest integer less than or equal to t. Then, $lim_{x \to 1^{+}} \frac{( 1 - ∣ x ∣ + s i n ∣1 - x ∣ ) s i n ( \frac{π}{2} [ 1 - x ])}{∣1 - x ∣ [ 1 - x ]}$

The value of $lim_{x \to 0^{+}} \frac{c o s ^{- 1} ( x - [ x ] ^{2} ) \cdot s i n ^{- 1} ( x - [ x ] ^{2} )}{x - x ^{3}}$ , where $[x]$ denotes the greatest integer $\leq x$ is :