If `f(x)={((tan^-1(x+[x]))/([x]-2x)[x]ne0,,),(0[x]=0,,):}` where `[x]` denotes the greatest integer less than or equal to x, than `lim_(xto0) f(x)` is (a) `(-1)/2` (b) 1 (c) `(pi)/4` (d) Does not exist

Options

A. `(-1)/2`
B. 1
C. `(pi)/4`
D. Does not exist

Answer

To solve the problem, we need to analyze the function \( f(x) \) given by:

\[ f(x) = \begin{cases} \frac{\tan^{-1}(x + [x])}{[x] - 2x} & \text{if } [x] \neq 0 \\ 0 & \text{if } [x] = 0 \end{cases} \]

where \([x]\) denotes the greatest integer less than or equal to \(x\).

We want to find the limit:

\[ \lim_{x \to 0} f(x) \]

### Step 1: Analyze the function around \( x = 0 \)

As \( x \) approaches \( 0 \), we need to consider both the left-hand limit (as \( x \to 0^{-} \)) and the right-hand limit (as \( x \to 0^{+} \)).

### Step 2: Calculate the left-hand limit \( \lim_{x \to 0^{-}} f(x) \)

For \( x \to 0^{-} \), we have \( [x] = -1 \) (since the greatest integer less than or equal to any negative number close to zero is -1). Thus, we can rewrite \( f(x) \):

\[ f(x) = \frac{\tan^{-1}(x - 1)}{-1 - 2x} \]

Now, we need to evaluate:

\[ \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} \frac{\tan^{-1}(x - 1)}{-1 - 2x} \]

As \( x \to 0^{-} \), \( \tan^{-1}(x - 1) \) approaches \( \tan^{-1}(-1) = -\frac{\pi}{4} \) and \( -1 - 2x \) approaches \(-1\).

Thus, we have:

\[ \lim_{x \to 0^{-}} f(x) = \frac{-\frac{\pi}{4}}{-1} = \frac{\pi}{4} \]

### Step 3: Calculate the right-hand limit \( \lim_{x \to 0^{+}} f(x) \)

For \( x \to 0^{+} \), we have \( [x] = 0 \). Thus, we can rewrite \( f(x) \):

\[ f(x) = \frac{\tan^{-1}(x + 0)}{0 - 2x} = \frac{\tan^{-1}(x)}{-2x} \]

Now, we need to evaluate:

\[ \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} \frac{\tan^{-1}(x)}{-2x} \]

As \( x \to 0^{+} \), \( \tan^{-1}(x) \) approaches \( 0 \). Therefore, we need to apply L'Hôpital's Rule since we have the indeterminate form \( \frac{0}{0} \):

\[ \lim_{x \to 0^{+}} \frac{\tan^{-1}(x)}{-2x} = \lim_{x \to 0^{+}} \frac{\frac{d}{dx}(\tan^{-1}(x))}{\frac{d}{dx}(-2x)} = \lim_{x \to 0^{+}} \frac{\frac{1}{1+x^2}}{-2} = \frac{1}{-2} = -\frac{1}{2} \]

### Step 4: Compare the left-hand and right-hand limits

We have:

- Left-hand limit: \( \lim_{x \to 0^{-}} f(x) = \frac{\pi}{4} \)
- Right-hand limit: \( \lim_{x \to 0^{+}} f(x) = -\frac{1}{2} \)

Since the left-hand limit and the right-hand limit are not equal, we conclude that:

\[ \lim_{x \to 0} f(x) \text{ does not exist.} \]

### Final Answer

The limit does not exist.