(A) √A+1 (B) √A (C) √A+5 (D) √A+21

Q: Wait, why do we use the Sandwich Theorem for the limit? Can't we just plug in zero?

Great question! If you try to plug in $x=0$ directly, you get $0 \times [\infty]$, which is an indeterminate form. The Greatest Integer Function is not continuous, so we cannot simply evaluate the limit by substitution. The Sandwich Theorem allows us to bound the expression $x[\frac{4}{x}]$ between $4-x$ and $4$, forcing the limit to be $4$ as $x$ approaches $0$.

Q: Why is the function continuous at integers like $x=2$?

At $x=2$, the function value is $f(2) = [4] \sin(2\pi) = 4 \times 0 = 0$. As $x \to 2^-$, $[x^2]$ approaches $3$, but $\sin(\pi x)$ approaches $0$, so the limit is $3 \times 0 = 0$. As $x \to 2^+$, $[x^2]$ approaches $4$, but $\sin(\pi x)$ approaches $0$, so the limit is $4 \times 0 = 0$. Since the left-hand limit, right-hand limit, and function value are all equal to $0$, it is continuous!

Q: Why is the function discontinuous at $x=\sqrt{5}$?

At $x=\sqrt{5}$, $x^2=5$. As $x \to \sqrt{5}^-$, $[x^2] \to 4$. As $x \to \sqrt{5}^+$, $[x^2] \to 5$. Crucially, $\sin(\pi \sqrt{5})$ is a non-zero constant. So, the left-hand limit is $4 \sin(\pi \sqrt{5})$ and the right-hand limit is $5 \sin(\pi \sqrt{5})$. Since $4 \neq 5$, the limits are unequal, and the function is discontinuous.

Question

Let $[t]$ denotes the greatest integer $\leq t$ and $\lim_{x 	o 0} x [\frac{4}{x}] = A$. Then the function, $f(x) = [x^2] \sin \pi x$ is discontinuous, when $x$ is equal to

Options:

(A) $\sqrt{A+1}$

(B) $\sqrt{A}$

(C) $\sqrt{A+5}$

(D) $\sqrt{A+21}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorers of mathematics! Today, we are going to unravel a problem that beautifully weaves together the rigid steps of the Greatest Integer Function and the smooth, oscillating waves of trigonometry.

This is a classic JEE Advanced challenge that tests not just your calculation skills, but your ability to visualize the behavior of functions at their breaking points.

### Taming the Limit

We begin with the limit $\lim_{x 	o 0} x [\frac{4}{x}] = A$. When you see a limit involving the Greatest Integer Function (GIF) as $x$ approaches zero, your first instinct should be the **Sandwich Theorem**.

The GIF is defined by the inequality $t-1 < [t] \le t$. By setting $t = \frac{4}{x}$, we get the inequality:

$$\frac{4}{x} - 1 < \left[ \frac{4}{x} ight] \le \frac{4}{x}$$

Now, we multiply this entire inequality by $x$. Assuming $x > 0$ for the right-hand limit, the inequality signs remain unchanged:

$$x \left( \frac{4}{x} - 1 ight) < x \left[ \frac{4}{x} ight] \le x \left( \frac{4}{x} ight)$$

This simplifies to $4 - x < x [\frac{4}{x}] \le 4$. As $x 	o 0$, the left side $4-x$ approaches $4$, and the right side is already $4$. By the **Sandwich Theorem**, the limit $A$ must be $4$.

### The Anatomy of Discontinuity

Now, let us examine the function $f(x) = [x^2] \sin(\pi x)$. This function is a product of two very different beasts.

The term $[x^2]$ is a step function that jumps whenever $x^2$ hits an integer. The term $\sin(\pi x)$ is a smooth, periodic wave.

A product of two functions is generally continuous where both are continuous. However, if $[x^2]$ has a jump discontinuity at $x=c$, the limit of $f(x)$ as $x 	o c$ will only exist if the other factor, $\sin(\pi x)$, is zero at $x=c$.

If $\sin(\pi c) = 0$, it acts as a 'mathematical sponge,' absorbing the jump and smoothing out the discontinuity. We know $\sin(\pi x) = 0$ whenever $x$ is an integer. Thus, at any integer $x$, the function is continuous!

### Finding the Break

The function is discontinuous at points where $x^2$ is an integer, but $x$ is NOT an integer. These are points like $x = \sqrt{2}, \sqrt{3}, \sqrt{5}, \dots$.

At these points, $[x^2]$ jumps, and since $x$ is not an integer, $\sin(\pi x)$ is not zero. The jump is not 'healed.'

Now, let us evaluate our options using $A=4$:

(A) $\sqrt{A+1} = \sqrt{5}$

(B) $\sqrt{A} = \sqrt{4} = 2$

(C) $\sqrt{A+5} = \sqrt{9} = 3$

(D) $\sqrt{A+21} = \sqrt{25} = 5$

Options (B), (C), and (D) result in integers ($2, 3, 5$). As we established, the function is continuous at integers.

Option (A) results in $\sqrt{5}$, which is not an integer. Therefore, the jump in $[x^2]$ at $x^2=5$ remains, and the function is **discontinuous at $x=\sqrt{5}$**.

### Conclusion

Mathematics is often about finding the balance between chaos and order. Here, the sine function provided the order, and the GIF provided the chaos.

By identifying where the order fails to contain the chaos, we found our answer. Keep practicing, keep visualizing, and most importantly, keep falling in love with the logic behind the math!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $[t]$ denotes the greatest integer $\leq t$ and $lim_{x \to 0} x [\frac{4}{x}] = A$ . Then the function, $f (x) = [x^{2}] sin π x$ is discontinuous, when $x$ is equal to

Select Answer:

Visualized Solution

The Problem Setup

Bounding the Greatest Integer Function

Setting up the Inequality

Applying the Sandwich Theorem

The Value of $A$

Analyzing the Function $f (x)$

The Smoothing Effect of Sine

Continuity at Integer Points

Discontinuity at Non-Integer Roots

Evaluating the Options

Final Conclusion

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

Taming the Limit

The Anatomy of Discontinuity

Finding the Break

Conclusion

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