Q.If $f(x) = \begin{cases} \dfrac{\sin[x]}{[x]}, & [x] \neq 0 \\ 0, & [x] = 0 \end{cases}$, where $[.]$ denotes the greatest integer function, then $\lim_{x \to 0} f(x)$ is equal to
(A) $1$
(B) $0$
(C) $-1$
(D) None of these

The key idea here is evaluating the limit of a function involving the greatest integer function by considering left-hand and right-hand limits.

1. To find $\lim_{x \to 0} f(x)$, we must evaluate the left-hand limit (LHL) and the right-hand limit (RHL) separately, as the greatest integer function $[x]$ changes its value around $x=0$.
2. For the LHL, as $x \to 0^-$, $x$ is slightly less than $0$ (e.g., $x = -0.001$). In this case, $[x] = -1$. Thus, $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin[x]}{[x]} = \frac{\sin(-1)}{-1} = \sin(1)$.
3. For the RHL, as $x \to 0^+$, $x$ is slightly greater than $0$ (e.g., $x = 0.001$). In this case, $[x] = 0$. …

When dealing with limits involving the greatest integer function, denoted by $[x]$, it is crucial to understand its behavior around integer points. The greatest integer function is discontinuous at every integer. This means that as $x$ approaches an integer from the left side, $[x]$ takes a different value than when $x$ approaches the same integer from the right side.

For a limit $\lim_{x \to a} f(x)$ to exist, the left-hand limit (LHL) and the right-hand limit (RHL) must both exist and be equal. This is especially important for functions defined piecewise or those involving components like $[x]$ that change their value abruptly.

In this problem, we need to find $\lim_{x \to 0} f(x)$. Since $x=0$ is an integer, we must evaluate the LHL and RHL separately.

1. Understand the function definition: The function $f(x)$ is defined as:

We need to determine the value of $[x]$ as $x$ approaches $0$ from the right and from the left.

2. Evaluate the Right-Hand Limit (RHL):

We consider $x \to 0^+$. This means $x$ is a very small positive number, for example, $x = 0.001$.

For any $x$ such that $0 < x < 1$, the greatest integer $[x]$ is $0$.

Since $[x] = 0$ for $x \in (0, 1)$, we use the second part of the function definition, which states $f(x) = 0$ when $[x] = 0$.

Therefore, for $x \to 0^+$, $f(x)$ is simply $0$.

3. Evaluate the Left-Hand Limit (LHL): We consider $x \to 0^-$. This means $x$ is a very small negative number, for example, $x = -0.001$. For any $x$ such that $-1 < x < 0$, the greatest integer $[x]$ is $-1$. Since $[x] = -1$, which is not equal to $0$, we use the first part of the function definition: $f(x) = \dfrac{\sin[x]}{[x]}$. Substitute $[x] = -1$ into this expression: …