`lim _(xto 1 ^(-)) (e ^({x}) - {x} -1)/( {x}^(2))` equal, where {.} is fractional part function and I is aan integer, to :

Options

A. `I/2`
B. `e -2`
C. `I `
D. does not exist

Answer

To solve the limit

\[ \lim_{x \to 1^-} \frac{e^{\{x\}} - \{x\} - 1}{\{x\}^2} \]

where \(\{x\}\) is the fractional part function, we can follow these steps:

### Step 1: Understand the Fractional Part Function
The fractional part function \(\{x\}\) is defined as:

\[ \{x\} = x - \lfloor x \rfloor \]

where \(\lfloor x \rfloor\) is the greatest integer less than or equal to \(x\). As \(x\) approaches 1 from the left, \(\lfloor x \rfloor = 0\), so:

\[ \{x\} = x - 0 = x \]

### Step 2: Substitute the Fractional Part
Substituting \(\{x\}\) into the limit gives us:

\[ \lim_{x \to 1^-} \frac{e^{x} - x - 1}{x^2} \]

### Step 3: Evaluate the Limit
Now we need to evaluate the limit as \(x\) approaches 1:

\[ \lim_{x \to 1^-} \frac{e^{x} - x - 1}{x^2} \]

Substituting \(x = 1\):

\[ \frac{e^{1} - 1 - 1}{1^2} = \frac{e - 2}{1} = e - 2 \]

### Conclusion
Thus, the limit evaluates to:

\[ \lim_{x \to 1^-} \frac{e^{\{x\}} - \{x\} - 1}{\{x\}^2} = e - 2 \]

### Final Answer
The final answer is:

\[ \boxed{e - 2} \]

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