Step 1: Understanding the Question:
This question is from the topic of limits and continuity in calculus.
We are given a piecewise-defined function that involves the greatest integer function $[x]$ and the fractional part function $\{x\}$. We need to check whether the limit of the function exists at $x = 0$ by evaluating the left-hand limit (LHL) and right-hand limit (RHL).
Step 2: Key Formulas and Approach:
Fractional Part Function Definition: $\{x\} = x - [x]$
Existence of a Limit: The limit exists if and only if:
\[
\text{LHL} = \text{RHL}
\]
Standard Limit: $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$
Step 3: Detailed Explanation:
Evaluate Right-Hand Limit (RHL) as $x \to 0^+$:
For $x \to 0^+$, $x$ is a small positive number ($0<x<1$).
Thus, $[x] = 0$ and $\{x\} = x - [x] = x - 0 = x$.
Substitute these values into the first branch of the function:
\[
\text{RHL} = \lim_{x \to 0^+} \frac{\tan^2 x}{x^2 - 0} = \lim_{x \to 0^+} \left(\frac{\tan x}{x}\right)^2 = (1)^2 = 1
\]
Evaluate Left-Hand Limit (LHL) as $x \to 0^-$:
For $x \to 0^-$, $x$ is a small negative number ($-1<x<0$).
Thus, $[x] = -1$ and $\{x\} = x - [x] = x - (-1) = x + 1$.
Substitute these values into the third branch of the function:
\[
\text{LHL} = \lim_{x \to 0^-} \sqrt{(x+1) \cot(x+1)}
\]
Substitute $x = 0$ directly into the continuous function:
\[
\text{LHL} = \sqrt{(0+1) \cot(0+1)} = \sqrt{1 \cdot \cot 1} = \sqrt{\cot 1}
\]
Compare LHL and RHL:
Since $\text{RHL} = 1$ and $\text{LHL} = \sqrt{\cot 1}$, we have $\text{LHL} \neq \text{RHL}$.
Therefore, the limit of the function does not exist at $x = 0$.
Step 4: Final Answer:
The limit of $f(x)$ as $x \to 0$ does not exist, which corresponds to Option (D).