Limit the fractional part

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

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

Here h a infinitesimally small positive number

So 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.

I just visualised the graph of the fractional part of x.

(One can consult ditutor for it.)

Then from the diagram if I move towards any integer from the left side(i.e. negative side) of the graph I reach to 1. So on putting {x} = 1 I got the answer e-2.