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Two problems - Limit and ContourPlot

Posted 10 years ago

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 Implicit Formula with ContourPlot[ x^(2/3) + y^(2/3) == 4, {x, -7, 7}, {y, -7, 7}]. It only gives the quarter of the graph!! Graph

POSTED BY: Ehsan Monabbati
6 Replies
Posted 10 years ago

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

enter image description here

POSTED BY: Jim Baldwin
Posted 10 years ago

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
POSTED BY: Jim Baldwin
Posted 10 years ago

Sorry! I have omitted the limit point: Limit[Sin[Sqrt[x^2 - 1]] - Sin[Sqrt[x^2 + 1]], x -> Infinity]. By using the command Plot[Sin[Sqrt[x^2 - 1]] - Sin[Sqrt[x^2 + 1]], {x, 0, 100}, PlotRange -> All] we can see the limit is equal to 0; but Mathematica 9 gives Interval[{-2,2}]!

enter image description here

POSTED BY: Ehsan Monabbati
Posted 10 years ago

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.

POSTED BY: Ehsan Monabbati
Posted 10 years ago

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.

POSTED BY: Jim Baldwin

Likewise for Surd[x, 3]^2 + Surd[y, 3]^2 == 4.

POSTED BY: Daniel Lichtblau
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