`undersetlim_(Xrarr2^(+)) {x}(sin(x-2))/((x-2)^(2))=` (where `{.}` denotes the fractional part function)

Options

A. 0
B. 2
C. 1
D. does not exist

Explanation

To solve the limit problem given, we need to evaluate:

\[ \lim_{x \to 2^+} \frac{\sin(x - 2)}{(x - 2)^2} \]

### Step 1: Substitute \( y = x - 2 \)

As \( x \to 2^+ \), \( y \to 0^+ \). Thus, we can rewrite the limit in terms of \( y \):

\[ \lim_{y \to 0^+} \frac{\sin(y)}{y^2} \]

### Step 2: Use the limit property of sine

We know that:

\[ \lim_{y \to 0} \frac{\sin(y)}{y} = 1 \]

This implies:

\[ \sin(y) \approx y \quad \text{as } y \to 0 \]

### Step 3: Rewrite the limit

Using the approximation, we can rewrite the limit as:

\[ \lim_{y \to 0^+} \frac{\sin(y)}{y^2} = \lim_{y \to 0^+} \frac{y}{y^2} = \lim_{y \to 0^+} \frac{1}{y} \]

### Step 4: Evaluate the limit

As \( y \to 0^+ \), \( \frac{1}{y} \) approaches \( +\infty \). Therefore, we have:

\[ \lim_{y \to 0^+} \frac{\sin(y)}{y^2} = +\infty \]

### Step 5: Determine the fractional part

Since the limit approaches \( +\infty \), we need to find the fractional part of \( +\infty \). The fractional part function, denoted by \( \{x\} \), is defined as \( x - \lfloor x \rfloor \). Since \( +\infty \) does not have a finite integer part, the fractional part does not exist.

### Conclusion

Thus, we conclude:

\[ \lim_{x \to 2^+} \left\{ \frac{\sin(x - 2)}{(x - 2)^2} \right\} \text{ does not exist.} \]

### Final Answer

The answer is that the limit does not exist.

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