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Calculate this Limit?

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?

21 Replies
Anonymous User
Anonymous User
Posted 8 years ago

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!

POSTED BY: Anonymous User

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.

POSTED BY: Gianluca Gorni

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.)

POSTED BY: Daniel Lichtblau

There are different counterexamples to L'Hopital's rule, e.g. in:

"Counterexamples to L'Hopital's rule"

Yes, if you don't care about the precise conditions, you find "counterexamples". The "counterexample" in the paper is very interesting, but it relies on infinitely many oscillations and algebraic cancellation, a combination that is unlikely to happen in the wild. I would say that with L'Hopital's rule you can most safely get away with ignoring the condition that the denominator be monotonic, unless somebody willfully laid a trap for you.

POSTED BY: Gianluca Gorni

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

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]]]}]
POSTED BY: Gianluca Gorni

Another strange result for the Limit function:

Limit[(Power[15 + 2 x, (3)^-1] + 1)/(Power[9 + x, (3)^-1] + x + 7), 
 x -> -8]

enter image description here

How to understand it?

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...

POSTED BY: Sander Huisman

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

enter image description here

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.

enter image description here

Answer from Maple

enter image description here

POSTED BY: Mariusz Iwaniuk

Thank you, Mariusz. Is this a bug? Must we report it to WRI?

It's no a Bug. It's seems a Simplify with helps ComplexExpand command simplifies and solves this problem.

POSTED BY: Mariusz Iwaniuk

Agree!

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]
POSTED BY: Gianluca Gorni

Thank you, Gianluca, for a simple and clear answer to the question!

Thank you, Gianluca!

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]
POSTED BY: Gianluca Gorni

Thank you, Mariusz?

It's NOT a error, because from Mathematica Documentation Center:

enter image description here

POSTED BY: Mariusz Iwaniuk
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