If `f(x)={{:((sin[x])/([x])","" ""for "[x]ne0),(0","" ""for "[x]=0):}`where `[x]` denotes the greatest integer less than or equal to x. Then find `lim_(xto0)f(x).`

Explanation

To find the limit of the function \( f(x) \) as \( x \) approaches 0, we analyze the function defined as follows:

\[ f(x) = \begin{cases} \frac{\sin([\!x])}{[\!x]} & \text{if } [\!x] \neq 0 \\ 0 & \text{if } [\!x] = 0 \end{cases} \]

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

### Step 1: Analyze the limit from the right (\( x \to 0^+ \))
When \( x \) approaches \( 0 \) from the right (i.e., \( x \to 0^+ \)), the greatest integer function \([\!x]\) will be \( 0 \). Therefore, we can substitute this into the function:

\[ f(x) = 0 \quad \text{for } [\!x] = 0 \]

Thus,

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

### Step 2: Analyze the limit from the left (\( x \to 0^- \))
When \( x \) approaches \( 0 \) from the left (i.e., \( x \to 0^- \)), the greatest integer function \([\!x]\) will be \( -1 \) (since \( x \) is slightly less than 0). Therefore, we substitute this into the function:

\[ f(x) = \frac{\sin(-1)}{-1} = \frac{-\sin(1)}{-1} = \sin(1) \]

Thus,

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

### Step 3: Compare the two limits
Now we have:

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

Since \( 0 \) is not equal to \( \sin(1) \), the two one-sided limits are not equal.

### Conclusion
Therefore, the limit \( \lim_{x \to 0} f(x) \) does not exist.

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

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