I used Mathematica to find a limit:
In[1]:= Limit[(Log[1 - x] - Sin[x])/(1 - Cos[x]^2), x -> 0]
Out[1]= -\[Infinity]
But the function (Log[1 - x] - Sin[x])/(1 - Cos[x]^2) doesn’t have the limit -[Infinity] in the point x=0.
It’s easy to observe that for x->+0 the limit is -[Infinity],
and for x->-0 the limit is +[Infinity].
The graph confirms this fact:
Plot[(Log[1 - x] - Sin[x])/(1 - Cos[x]^2), {x, -2, 1}]
Is this a Mathematica error?
It’s NOT a error, because from Mathematica Documentation Center:
Thank you, Mariusz?
By default Limit takes the limit from the right. The option Direction is there for us to choose the direction: -1 is for the limit from the right (in the negative direction) and 1 for the limit from the left (in the positive direction):
Limit[(Log[1 - x] - Sin[x])/(1 - Cos[x]^2), x -> 0, Direction -> 1]
Limit[(Log[1 - x] - Sin[x])/(1 - Cos[x]^2), x -> 0, Direction -> -1]
Thank you, Gianluca!
Another limit, the same difficulty…
In the following example I intentionally gives the result at the beginning because it’s not difficult to solve the problem analytically.
So, I tried to verify In Mathematica the result of this example
The Mathematica gives
In[1]:= Limit[\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(n\)]\(Abs[
\*SuperscriptBox[\(E\),
FractionBox[\(2. \[Pi]\ I\ k\), \(n\)]] -
\*SuperscriptBox[\(E\),
FractionBox[\(2 \[Pi]\ I\ \((k - 1)\)\), \(n\)]]]\)\),
n -> \[Infinity]]
Out[1]= Limit[\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(n\)]\(Abs[
\*SuperscriptBox[\(E\),
FractionBox[\(\((0.` + 6.283185307179586`\ I)\)\ k\), \(n\)]] -
\*SuperscriptBox[\(E\),
FractionBox[\(2\ I\ \((\(-1\) + k)\)\ \[Pi]\), \(n\)]]]\)\),
n -> \[Infinity]]
It’s interesting that when I tried to compute simply the numerical evaluations of the expression under the limit, I obtained
In[2]:= Table[\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(n\)]\(Abs[
\*SuperscriptBox[\(E\),
FractionBox[\(2. \[Pi]\ I\ k\), \(n\)]] -
\*SuperscriptBox[\(E\),
FractionBox[\(2 \[Pi]\ I\ \((k - 1)\)\), \(n\)]]]\)\), {n, 10000,
10010}]
Out[2]= {6.28319, 6.28319, 6.28319, 6.28319, 6.28319, 6.28319, \
6.28319, 6.28319, 6.28319, 6.28319, 6.28319}
In[3]:= Table[\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(n\)]\(Abs[
\*SuperscriptBox[\(E\),
FractionBox[\(2. \[Pi]\ I\ k\), \(n\)]] -
\*SuperscriptBox[\(E\),
FractionBox[\(2 \[Pi]\ I\ \((k - 1)\)\), \(n\)]]]\)\), {n, 100000,
100010}]
Out[3]= {6.28319, 6.28319, 6.28319, 6.28319, 6.28319, 6.28319, \
6.28319, 6.28319, 6.28319, 6.28319, 6.28319}
These results are approximately equal to 2\pi. Why didn’t work the function Limit[]?
Problem is in the Sum, not in Limit.
Mathematica can’t find symbolic solution in Sum.
Answer from Maple
Thank you, Mariusz. Is this a bug? Must we report it to WRI?
Mathematica can do it with ComplexExpand:
Sum[Simplify@
ComplexExpand@Abs[Exp[2 Pi I k/n] - Exp[2 Pi I (k - 1)/n]], {k, n}]
Limit[%, n -> Infinity]
Thank you, Gianluca, for a simple and clear answer to the question!
It’s no a Bug. It’s seems a Simplify with helps ComplexExpand command simplifies and solves this problem.
Agree!
Basically it blows up to ‘an infinity’; not negative infinity, not positive infinite, or i*infinity, but roughly (0.866025 + 0.5 I) * infinity …
Check the FullForm of your expression to better understand it perhaps…
It is a DirectedInfinity. Your function is going to infinity in the complex plane along that direction. Check out the plot:
ParametricPlot[
ReIm[((15 + 2 x)^(1/3) + 1)/((9 + x)^(1/3) + x + 7)], {x, -8, -7.5},
AxesOrigin -> {0, 0},
Epilog -> {Dashed, HalfLine[{0, 0}, ReIm[(1 + (-1)^(1/3))/Sqrt[3]]]}]
Bingo! Finally, I solved it after clarifying that Mathematica has two different cube roots for Power[ ] function and Surd[ ] function. In the above example, It must be used Surd[ ] function. So:
In[1]:= Limit[(Surd[15 + 2 x, 3] + 1)/(Surd[9 + x, 3] + x + 7),
x -> -8]
Out[1]= 1/2
btw L’Hopital rule is flawed i hope your not using that to determine if Mathematica is correct
your comparing points on two function lines, but l’hopital tries a trick using one line, the data isn’t always there: L’Hopital’s rules is plain wrong - wrong but easy!
The task was to compute directly the limit with the Mma standard function. So, we found that Surd[] is more appropriate in this case that Power[].
Or, we can use CubeRoot[]:
In[1]:= Limit[(CubeRoot[15 + 2 x] + 1)/(CubeRoot[9 + x] + x + 7),
x -> -8]
Out[1]= 1/2
The L’Hopital rule that I know is very correct. Perhaps you have another rule in mind.
To add to the remark by @GianlucaGorni, l’Hopital’s rule gives the correct result here.
D[(CubeRoot[15 + 2 x] + 1), x]/D[CubeRoot[9 + x] + x + 7, x] /. x -> -8
(* Out[960]= 1/2 *)
This also raises a question: What exactly is “wrong” with l’Hopital’s rule? (There is some literature on this, by the way.)