Let [x] denote the greatest integer less than or equal to x. Then lim x-0 (tan(π sin²x) + (|x| - si

Q: Why can't we just use L'Hopital's Rule directly on the whole expression?

L'Hopital's Rule is a powerful tool, but it assumes the function is differentiable and continuous in the neighborhood of the point. Because $[x]$ is a step function, it is not continuous at $x=0$. Differentiating a step function is mathematically undefined in the standard sense. Always check for continuity first!

Q: Why does $[x]$ become $-1$ when $x \to 0^-$?

The Greatest Integer Function $[x]$ returns the largest integer less than or equal to $x$. If $x$ is a tiny negative number like $-0.001$, the largest integer less than or equal to $-0.001$ is $-1$. It is a common mistake to think it becomes $0$.

Q: Is it always necessary to split the limit?

In JEE Advanced, whenever you encounter piecewise functions like $[x]$, $|x|$, or $\text{sgn}(x)$, splitting the limit is not just a suggestion; it is a necessity. It is the only way to rigorously prove the existence or non-existence of the limit.

Question

Let $[x]$ denote the greatest integer less than or equal to $x$. Then $\lim_{x 	o 0} \frac{	an(\pi \sin^2 x) + (|x| - \sin(x[x]))^2}{x^2}$ :

Options:

(A) equals $\pi$

(B) equals 0

(C) equals $\pi + 1$

(D) does not exist

Sigma Percentile · Accepted Answer

Welcome, student. Today we tackle a problem that separates the casual learners from the true masters of calculus. When you see the Greatest Integer Function $[x]$ and the Modulus function $|x|$ in a limit problem, you are not just looking at algebra; you are looking at a landscape that changes its terrain the moment you cross the integer boundary.

The problem asks us to evaluate:

$$\lim_{x 	o 0} \frac{	an(\pi \sin^2 x) + (|x| - \sin(x[x]))^2}{x^2}$$

The first instinct of a novice is to plug in $x=0$ and hope for the best. But we are JEE aspirants; we know that limits are about the journey, not just the destination. We must investigate the behavior from both sides.

### Analyzing the Right-Hand Limit (RHL)

Let us start with the Right-Hand Limit (RHL), where $x 	o 0^+$. In this region, $x$ is a tiny positive value. Consequently, the Greatest Integer Function $[x]$ is $0$, and the modulus $|x|$ is simply $x$.

Substituting these, our expression transforms into:

$$\frac{	an(\pi \sin^2 x) + (x - \sin(0))^2}{x^2} = \frac{	an(\pi \sin^2 x) + x^2}{x^2}$$

We can split this into two parts:

$$\frac{	an(\pi \sin^2 x)}{x^2} + 1$$

To solve the first part, we use the standard limit $\lim_{	heta 	o 0} \frac{	an 	heta}{	heta} = 1$. By multiplying and dividing by $\pi \sin^2 x$, we get:

$$\frac{	an(\pi \sin^2 x)}{\pi \sin^2 x} \cdot \frac{\pi \sin^2 x}{x^2} + 1$$

Since $\frac{\sin x}{x} 	o 1$, this evaluates to $\pi + 1$.

### Analyzing the Left-Hand Limit (LHL)

Now, let us pivot to the Left-Hand Limit (LHL), where $x 	o 0^-$. Here, $x$ is a tiny negative number. The Greatest Integer Function $[x]$ becomes $-1$, and the modulus $|x|$ becomes $-x$.

Substituting these, we get:

$$\frac{	an(\pi \sin^2 x) + (-x - \sin(-x))^2}{x^2}$$

Since $\sin(-x) = -\sin x$, the term inside the square becomes $(-x + \sin x)^2$, which is equivalent to $(x - \sin x)^2$. The expression becomes:

$$\frac{	an(\pi \sin^2 x)}{x^2} + \frac{(x - \sin x)^2}{x^2}$$

The first term remains $\pi$. The second term is:

$$\left( \frac{x - \sin x}{x} ight)^2 = \left( 1 - \frac{\sin x}{x} ight)^2$$

As $x 	o 0$, this becomes $(1 - 1)^2 = 0$. Thus, the LHL is $\pi + 0 = \pi$.

### The Verdict

Comparing our results, the RHL is $\pi + 1$ and the LHL is $\pi$. Because they are not equal, the **limit does not exist**.

This problem teaches us that in the world of JEE, precision is everything. Never assume continuity where there is none. Keep practicing, and remember: the math is always elegant if you look closely enough.

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

Select Answer:

Visualized Solution

Introduction to the Limit

Analyzing $x \to 0^{+}$ (RHL)

Substituting in RHL

Simplifying the RHL Expression

Splitting the RHL Limit

Evaluating the Tangent Term

Final RHL Value

Analyzing $x \to 0^{-}$ (LHL)

Substituting in LHL

Simplifying the LHL Expression

Splitting the LHL Limit

Evaluating LHL Terms

Conclusion and Key Takeaway

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

Analyzing the Right-Hand Limit (RHL)

Analyzing the Left-Hand Limit (LHL)

The Verdict

Similar Questions

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

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

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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

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