Compute $\lim _{ x \rightarrow { { n }^{ - }} }{ \frac { { e }^{\{ x \} }-\{ x \} -1 }{ { \{ x \} }^{ 2 } } }$

Where $\{.\}$ represents the fractional part function and $n$ is an integer.

1/2
e-2
I- The Integer
*Does not exist

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Here the $Limit$ $is$ $x \rightarrow I^-$

We can also write it as $x \rightarrow I - h$

Here h a infinitesimally small positive numberSo the expression becomes \lim_{h \to 0} \frac{e^\left\{I - h\right\} - \left\{I - h \right\} - 1}{\left\{I-h\right\}^2}

We replaced x by I - h where $h \rightarrow 0$Now as we are aware of the property $\left\{x\right\} = x - \lfloor x \rfloor$

So following this property, we have $\left\{I-h\right\} = (I-h) - \lfloor I-h \rfloor$

$\Rightarrow \left\{I-h\right\} = (I-h) - (I-1)$

as the greatest integer less than $I$ is $I-1$So we get $\Rightarrow \left\{I-h\right\} = 1-h$

Plugging in the value of $\left\{I-h\right\}$ into the main expression

$\Rightarrow \lim_{h \to 0} \frac{e^{(1-h)} - (1-h) - 1}{(1-h)^2}$

Put $h=0$ into the above expression and we get

$\frac{e^{(1-0)} - (1-0) - 1}{(1-0)^2}$

$\frac{e^{(1)} - 2}{(1)^2}$

$\Rightarrow e - 2$

Hence $e - 2$ is the correct answer.