Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $[t]$ denote the greatest integer $\le t$ and ${\lim }_{x\to 0}x\left[\frac{4}{x}\right]=A$. Then the function, $f(x)=[x^2]\sin (\pi x)$ is discontinuous, when $x$ is equal to :

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dissect a problem that perfectly illustrates the delicate balance between chaos and order in calculus.

We are dealing with the greatest integer function, $[t]$, a function that loves to jump, and the sine function, $\sin (\pi x)$, which loves to oscillate. When they meet, we find a fascinating study of continuity.

We begin with the expression $A={\lim }_{x\to 0}x\left[\frac{4}{x}\right]$. At first glance, this looks intimidating because as $x$ approaches zero, $\frac{4}{x}$ shoots off to infinity.

To resolve this, we use the identity $[t]=t-\{t\}$, where $\{t\}$ is the fractional part, trapped forever in the interval $[0,1)$. By substituting this into our limit, we get:

Here is the beauty of the Squeeze Theorem: even though $\{4/x\}$ is dancing erratically, it is bounded. When we multiply it by $x$, which is shrinking to zero, the entire term $x\{4/x\}$ is forced to zero.

Thus, $A=4-0=4$. We have successfully tamed the beast.

Now, we turn our attention to $f(x)=[x^2]\sin (\pi x)$. We know that $[x^2]$ is a step function that remains constant between integers, but at every point where $x^2$ is an integer, it takes a leap.

These are our candidates for discontinuity. Specifically, $[x^2]$ is discontinuous at $x=\sqrt{k}$ for any integer $k$.

However, we are multiplying this by $\sin (\pi x)$. Think of $\sin (\pi x)$ as a "healing factor." If $\sin (\pi x)$ happens to be zero at the exact point where $[x^2]$ jumps, the product might just be continuous.

For example, if $x=2$, then $x^2=4$. The function $[x^2]$ jumps at $x=2$, but $\sin (2\pi )=0$. The zero "annihilates" the jump, making the function continuous at that point.

We have $A=4$. Let us test our options to see where the discontinuity survives:

1. Option A ($x=\sqrt{4}=2$): As we discussed, $\sin (2\pi )=0$. The function is continuous here.

2. Option B ($x=\sqrt{4+1}=\sqrt{5}$): Here, $x^2=5$. The function $[x^2]$ jumps at $x=\sqrt{5}$.

Since $\sqrt{5}$ is irrational, $\sin(\pi \sqrt{5}) eq 0$. The sine term is non-zero, so it cannot save the discontinuity. The function is discontinuous at $x=\sqrt{5}$.

3. Option C ($x=\sqrt{4+5}=3$): Here, $x^2=9$. The sine term is $\sin (3\pi )=0$. The function is continuous here.

4. Option D ($x=\sqrt{4+21}=5$): Here, $x^2=25$. The sine term is $\sin (5\pi )=0$. The function is continuous here.

Through this journey, we have seen how a function can be "saved" from discontinuity by the presence of a zero-crossing in its multiplier.

The point $x=\sqrt{5}$ stands out as the only one where the jump of $[x^2]$ is left exposed. This is the elegance of JEE-level calculus: it is not just about calculation; it is about understanding the interaction between different mathematical structures.