If `f(x)={((sin([x]+x))/([x]+x),x!=0),(1,x=0):}` when [.] denotes the greatest integer function, then:

Options

A. `lim_(xto0)(f(x)=sin1`
B. `lim_(xto0)f(x)=1`
C. `lim_(xto0)f(x)` does not exist
D. `lim_(xto0)f(x)` exists but `f(x)` is not continuous at x=0

Answer

To determine the continuity of the function

\[ f(x) = \begin{cases} \frac{\sin([x] + x)}{[x] + x}, & x \neq 0 \\ 1, & x = 0 \end{cases} \]

where \([x]\) denotes the greatest integer function, we need to analyze the limits of \(f(x)\) as \(x\) approaches 0 from both the left and the right.

### Step 1: Calculate the Left-Hand Limit as \(x \to 0^-\)

For \(x < 0\), \([x] = -1\). Thus, we have:

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

Now, we find the left-hand limit:

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

Substituting \(x = 0\):

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

### Step 2: Calculate the Right-Hand Limit as \(x \to 0^+\)

For \(x \geq 0\), \([x] = 0\). Thus, we have:

\[ f(x) = \frac{\sin(0 + x)}{0 + x} = \frac{\sin(x)}{x} \]

Now, we find the right-hand limit:

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

Using the standard limit:

\[ = 1 \]

### Step 3: Check Continuity at \(x = 0\)

We know:

- Left-hand limit as \(x \to 0^- = \sin(1)\)
- Right-hand limit as \(x \to 0^+ = 1\)
- \(f(0) = 1\)

Since the left-hand limit \(\sin(1)\) is not equal to the right-hand limit \(1\), the limit as \(x\) approaches \(0\) does not exist.

### Conclusion

Since the left-hand limit and right-hand limit are not equal, the function \(f(x)\) is discontinuous at \(x = 0\).

### Final Answer

The limit does not exist at \(x = 0\), and thus the function is discontinuous at that point.

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