I tried evaluating the following limit with WA, but I got a wrong answer. The limit is: $$\lim_{n\to\infty}\left| \frac{n\sin^2(n+1)}{(n+1)\sin^2(n)}\right|$$ The result that I obtained from WA was $0$, however it's wrong. Any thoughts? How do I report this?

You can check this by:

    "Limit Abs[n Csc(n)^2 Sin(n+1)^2/(n+1)] as n goes to infinity"


$$$$ **Edit: Mathematica is not able evaluate the limit, instead other computing programs like Maple and Matlab seem to do so. Why so?

Why is the equivalent expression

Limit[ Abs[ n Csc[n]^2 Sin[n + 1]^2/(n + 1) ], n -> ∞]

returned unevaluated?

  • $\begingroup$ I would encourage you to fill out the feedback form at the bottom of the page, so the original input query is tagged with your comment. $\endgroup$ – Jason B. Mar 2 '17 at 5:21
  • $\begingroup$ @JasonB. Do you have any ideas why WolframAlpha gives this output? $\endgroup$ – user372003 Mar 2 '17 at 5:31
  • $\begingroup$ @user372003 What is the expected output? $\endgroup$ – zhk Mar 2 '17 at 5:50
  • $\begingroup$ @MMM the limit does not equal 0... you can clearly see from the graph... however, the expected output should be that it does not exist $\endgroup$ – user372003 Mar 2 '17 at 5:54
  • 1
    $\begingroup$ @gwr - I deleted the auto-generated comment from the vote, before I saw your reply. I just meant that the best way to get things fixed on alpha is to have it logged by the alpha developers, and the best way to do that is to press the feedback button on the page with a wrong answer. If this can be rewritten to focus on failure in Mathematica then it should be. $\endgroup$ – Jason B. Mar 2 '17 at 14:49

I may not be Mathematician enough, but one can clearly see that the function given is a periodic function:

f = Function[ n, Abs[(n Sin[n + 1]^2)/((n + 1) Sin[n]^2)]];

Plot[ f[n], {n, 0, 10}, ImageSize -> Large, PlotTheme -> "Scientific" ]


For every multiple of $\pi$ the function will have a pole:

Limit[ f[n], n -> π] // Simplify


And quite obviously for every value $ k\cdot\pi - 1, k \in \mathbb{Z} $ the function hits zero.

A more formal approach is discussed on Math.SE here: Accordingly choosing $x_n = n\cdot \pi $ and $y_n = n\cdot \pi - 1$ will both go to $\infty$ as $n \rightarrow \infty$, but clearly then $f(x_n) \neq f(y_n)$ so there cannot be a limit.


Take a look at $f(x) = |\csc(x)|$

Plot[ Abs[ Csc[x] ], {x,0,4 \[Pi] },  ]


and that function will give an interval for the Limit:

Limit[ Abs[ Csc[x] ], x -> Infinity ]


But $f(x) = \csc^2(x)$ will not and simply remains unevaluated:

Plot[ Csc[x]^2, {x, 0, 4 \[Pi]} ]


Limit[ Csc[x]^2, x -> Infinity ]

(* remains unevaluated *)

While you would expect to receive the exact same result as before, wouldn't you?

  • $\begingroup$ I just edited my answer because it does look inconsistent, how Limitwill treat Csc[x] and Csc[x]^2 or does it not? $\endgroup$ – gwr Mar 2 '17 at 14:30

Does this function have a limit? Let

f[n_] := Abs[n Sin[n + 1]^2/((n + 1) Sin[n]^2)] // N

Here we plot a few values:

ListPlot[f[#] & /@ Range[100000, 100200]]

enter image description here

It sure doesn't look like it has a limit...

  • $\begingroup$ In either case Wolfram Alpha is wrong? $\endgroup$ – user372003 Mar 2 '17 at 4:49



enter image description here



  • $\begingroup$ How about Mathematica? It does not give 0 as an output, however it does not evaluate the limit either $\endgroup$ – user372003 Mar 2 '17 at 7:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.