If {x} denotes the fractional part of x, then `lim_(x to 0) ({x})/(tan {x})` is equal to

Options

A. 1
B. 0
C. `-1`
D. Limit does not exist

Answer

To solve the limit problem \( \lim_{x \to 0} \frac{\{x\}}{\tan \{x\}} \), where \(\{x\}\) denotes the fractional part of \(x\), we will break down the solution step by step.

### Step 1: Understanding the Fractional Part
The fractional part of \(x\) is defined as:
\[ \{x\} = x - \lfloor x \rfloor \]
where \(\lfloor x \rfloor\) is the greatest integer less than or equal to \(x\).

### Step 2: Rewrite the Limit
We can rewrite the limit as:
\[ \lim_{x \to 0} \frac{x - \lfloor x \rfloor}{\tan(x - \lfloor x \rfloor)} \]

### Step 3: Evaluate Right-Hand Limit (RHL)
We first evaluate the right-hand limit as \(x\) approaches \(0\) from the positive side (\(0^+\)):
- As \(x \to 0^+\), \(\lfloor x \rfloor = 0\).
- Therefore, \(\{x\} = x - 0 = x\).

The limit becomes:
\[ \lim_{x \to 0^+} \frac{x}{\tan x} \]

Using the known limit:
\[ \lim_{x \to 0} \frac{\tan x}{x} = 1 \implies \lim_{x \to 0} \frac{x}{\tan x} = 1 \]

Thus, the right-hand limit is:
\[ \text{RHL} = 1 \]

### Step 4: Evaluate Left-Hand Limit (LHL)
Next, we evaluate the left-hand limit as \(x\) approaches \(0\) from the negative side (\(0^-\)):
- As \(x \to 0^-\), \(\lfloor x \rfloor = -1\).
- Therefore, \(\{x\} = x - (-1) = x + 1\).

The limit becomes:
\[ \lim_{x \to 0^-} \frac{x + 1}{\tan(x + 1)} \]

As \(x \to 0^-\), \(x + 1 \to 1\), so we need to evaluate:
\[ \lim_{x \to 0^-} \frac{1}{\tan(1)} \]

Thus, the left-hand limit is:
\[ \text{LHL} = \frac{1}{\tan(1)} \]

### Step 5: Compare RHL and LHL
Now, we compare the right-hand limit and the left-hand limit:
- RHL = 1
- LHL = \frac{1}{\tan(1)}

Since \(1 \neq \frac{1}{\tan(1)}\), the limits do not match.

### Conclusion
Since the right-hand limit does not equal the left-hand limit, the limit does not exist:
\[ \lim_{x \to 0} \frac{\{x\}}{\tan \{x\}} \text{ does not exist.} \]