In Mathematica 8 the code Limit[ Sin[Sqrt[x^2-1]]-Sin[Sqrt[x^2+1]]] gives the correct answer 0 but Mathematica 9 and WolframAlpha both give Interval[{-2,2}], what is wrong here??
Another problem I encountered is in plotting implicit relation
with ContourPlot[ x^(2/3) + y^(2/3) == 4, {x, -7, 7}, {y, -7, 7}]. It only gives the quarter of the graph!!
Looks like something is missing from the Limit function:
Limit[Sin[Sqrt[x^2 - 1]] - Sin[Sqrt[x^2 + 1]], x -> x0]
where x0 is set to the desired value. The function in question also has multiple values of x0 that result in a limit/value of zero (although it doesn’t appear that it’s necessary to use Limit - but I must be missing something).
Plot[Sin[Sqrt[x^2 - 1]] - Sin[Sqrt[x^2 + 1]], {x, -20, 20},
PlotRange -> {{-20, 20}, {-1, 0.5}}]
results in
About the contour plot…I think that the following explains the result:
x^(2/3) /. x -> -7
N[x^(2/3) /. x -> -7]
(x^2)^(1/3) /. x -> -7
N[(x^2)^(1/3) /. x -> -7]
with corresponding results
(-7)^(2/3)
-1.82965 + 3.16905 I
7^(2/3)
3.65931
Thank you for your very nice point!
Now, the command
ContourPlot[ (x^2)^(1/3) + (y^2)^(1/3) == 4, {x, -7, 7}, {y,-7,7} ]
gives the correct answer.
Uh, this isn’t the kind of answer I like to get but for Mathematica 10.0.1 (Windows 7 and a more recent version than 8 and 9 that you mention), I get the correct answer:
In[1]:= Limit[Sin[Sqrt[x^2 - 1]] - Sin[Sqrt[x^2 + 1]], x -> Infinity]
Out[1]= 0
And for whatever it’s worth I get the same response as you for version 9.
Likewise for Surd[x, 3]^2 + Surd[y, 3]^2 == 4.