I suppose this is due to rounding. Since i can't comment on posts yet, here's what i have found.

Given x*i+1-1

For, i<0.5*10^15 $MachineEpsilon I get zero.

While for i>0.5*10^15 $MachineEpsilon i get $MachineEpsilon.

And regarding your second question, on my machine

$MachineEpsilon = 2.22045×10−16

It might be a naive way to show it but here's an illustration.

DiscretePlot[{(x*i) + 1 - 1}, {i, 1.1101, 1.11035, 0.000001}]

I suppose this is due to rounding. For as long as the value is smaller than half of $MachineEpsilon it yields 0, else it rounds up to $MachineEpsilon.

Given x*i+1-1

For, i<0.5*10^15 $MachineEpsilon I get zero,
while for i>0.5*10^15 $MachineEpsilon i get $MachineEpsilon.

And regarding your second question: On my laptop

$MachineEpsilon = 2.22045×10−16

It might be a naive way to show it, but here's an illustration.

DiscretePlot[{(x*i) + 1 - 1}, {i, 1.1101, 1.11035, 0.000001}]

I suppose this is due to rounding. Since i can't comment on posts yet, here's what i have found.

Given x*i+1-1

For, i<0.5*$MachineEpsilon I get zero.

While for i>0.5*$MachineEpsilon i get $MachineEpsilon.

And regarding your second question, on my machine

$MachineEpsilon = 2.22045×10−16

I suppose this is due to rounding. Since i can't comment on posts yet, here's what i have found.

Given x*i+1-1

For, i<0.5*10^15 $MachineEpsilon I get zero.

While for i>0.5*10^15 $MachineEpsilon i get $MachineEpsilon.

And regarding your second question, on my machine

$MachineEpsilon = 2.22045×10−16

It might be a naive way to show it but here's an illustration.

DiscretePlot[{(x*i) + 1 - 1}, {i, 1.1101, 1.11035, 0.000001}]