Comment on Two Poll Questions

One Question

One of the super interesting Poll questions today was about finding the limit of
Limit
x->n
x
IntegerPart[x]
We can let Mathematica do it:
In[]:=
Limit[x/IntegerPart[x],x->n]
Out[]=
n
IntegerPart[n]
Indeed, the limit seems to exist as the directional limits are equal to each other:
In[]:=
Limit[x/IntegerPart[x],x->n,Direction->"FromAbove"]
Out[]=
n
IntegerPart[n]
In[]:=
Limit[x/IntegerPart[x],x->n,Direction->"FromBelow"]
Out[]=
n
IntegerPart[n]
Let’s be more specific then, and let’s tell Mathematica that we want n to be integer. Curiously, the result is:
In[]:=
Assuming[n∈Integers,Limit[x/IntegerPart[x],x->n]]
Out[]=
Indeterminate
The question is WHY? And the answer comes from the fact that zero is an integer (and so is -1). Indeed,
In[]:=
0∈Integers
Out[]=
True
Thus for only one integer,
n=0
, we have the incongruity of the directional limits. Indeed, for
0<n<1
,
n
IntegerPart[n]
=
n
0
, which is indeterminate, whereas for
-1<n<0
,
n
IntegerPart[n]
=
n
-1
, is determinate.
How about checking the directional limits with
x->0
?
In[]:=
Limit[x/IntegerPart[x],x->0,Direction->"FromAbove"]
Out[]=
ComplexInfinity
In[]:=
Limit[x/IntegerPart[x],x->0,Direction->"FromBelow"]
Out[]=
ComplexInfinity
So it would seem that the directional limits are equal to each other, doesn’t it?
In[]:=
Limit[x/IntegerPart[x],x->0,Direction->"FromAbove"]==Limit[x/IntegerPart[x],x->0,Direction->"FromBelow"]
Out[]=
ComplexInfinityComplexInfinity
Actually, NO! Because infinity is not necessarily equal to infinity. The answer to the equation:
ComplexInfinityComplexInfinity
should be
Indeterminate
. It’s curious that Mathematica refuses to make this evaluation.
This notwithstanding, THANK YOU, Luke, for this very interesting problem.

Second Question

Another question seemed to imply that the functions
Sin[x]
x
and
x
Sin[x]
are discontinuous. They are not; they are continuous. Indeed, considering the first one, we have:
In[]:=
Limit
Sin[x]
x
,x->0,​​Limit
Sin[x]
x
,x->0,Direction->"FromAbove",​​Limit
Sin[x]
x
,x->0,Direction->"FromBelow"
Out[]=
{1,1,1}
Furthermore, Taylor expansion around the “problem point” is a simple polynomial:
In[]:=
Series
Sin[x]
x
,{x,0,5}
Out[]=
1-
2
x
6
+
4
x
120
+
6
O[x]