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Why wolfram returns a bad response to a limit?

Posted 10 years ago

A friend asked me about a limit, use it to check your answer wolfram but ... wolfram returns another value

lim (x to -infinity ) ((x+5sqr(4+x²))/((9+8x³)^(1/3)))
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POSTED BY: Eduen Sarceno
6 Replies

Hi,

I still think that your calculations incorrect at these places. If $x\rightarrow -\infty$ the nominator is real and the denominator is complex. The whole thing will go complex. Also look a this:

Table[Evaluate[(x + 5 Sqrt[4 + x^2])/((9 + 8 x^3)^(1/3))] // N, {x, 0, -100, -1}]

{4.8075, 10.1803, 1.59641 - 2.76506 I, 1.27021 - 2.20006 I, 
 1.15435 - 1.99939 I, 1.0996 - 1.90456 I, 1.06948 - 1.85239 I, 
 1.05117 - 1.82068 I, 1.03923 - 1.8 I, 1.03102 - 1.78578 I, 
 1.02514 - 1.77559 I, 1.02078 - 1.76804 I, 1.01746 - 1.7623 I, 
 1.01488 - 1.75782 I, 1.01283 - 1.75427 I, 1.01117 - 1.75141 I, 
 1.00982 - 1.74906 I, 1.0087 - 1.74712 I, 1.00776 - 1.74549 I, 
 1.00696 - 1.74411 I, 1.00628 - 1.74293 I, 1.0057 - 1.74192 I, 
 1.00519 - 1.74104 I, 1.00475 - 1.74027 I, 1.00436 - 1.7396 I, 
 1.00402 - 1.73901 I, 1.00371 - 1.73848 I, 1.00344 - 1.73802 I, 
 1.0032 - 1.7376 I, 1.00298 - 1.73722 I, 1.00279 - 1.73688 I, 
 1.00261 - 1.73657 I, 1.00245 - 1.7363 I, 1.0023 - 1.73604 I, 
 1.00217 - 1.73581 I, 1.00205 - 1.7356 I, 1.00194 - 1.7354 I, 
 1.00183 - 1.73522 I, 1.00174 - 1.73506 I, 1.00165 - 1.73491 I, 
 1.00157 - 1.73477 I, 1.00149 - 1.73463 I, 1.00142 - 1.73451 I, 
 1.00136 - 1.7344 I, 1.0013 - 1.73429 I, 1.00124 - 1.7342 I, 
 1.00118 - 1.7341 I, 1.00113 - 1.73402 I, 1.00109 - 1.73394 I, 
 1.00104 - 1.73386 I, 1.001 - 1.73379 I, 1.00096 - 1.73372 I, 
 1.00093 - 1.73366 I, 1.00089 - 1.7336 I, 1.00086 - 1.73354 I, 
 1.00083 - 1.73349 I, 1.0008 - 1.73343 I, 1.00077 - 1.73339 I, 
 1.00074 - 1.73334 I, 1.00072 - 1.7333 I, 1.0007 - 1.73326 I, 
 1.00067 - 1.73322 I, 1.00065 - 1.73318 I, 1.00063 - 1.73314 I, 
 1.00061 - 1.73311 I, 1.00059 - 1.73308 I, 1.00058 - 1.73305 I, 
 1.00056 - 1.73302 I, 1.00054 - 1.73299 I, 1.00053 - 1.73296 I, 
 1.00051 - 1.73294 I, 1.0005 - 1.73291 I, 1.00048 - 1.73289 I, 
 1.00047 - 1.73286 I, 1.00046 - 1.73284 I, 1.00045 - 1.73282 I, 
 1.00043 - 1.7328 I, 1.00042 - 1.73278 I, 1.00041 - 1.73276 I, 
 1.0004 - 1.73275 I, 1.00039 - 1.73273 I, 1.00038 - 1.73271 I, 
 1.00037 - 1.7327 I, 1.00036 - 1.73268 I, 1.00035 - 1.73267 I, 
 1.00035 - 1.73265 I, 1.00034 - 1.73264 I, 1.00033 - 1.73262 I, 
 1.00032 - 1.73261 I, 1.00032 - 1.7326 I, 1.00031 - 1.73259 I, 
 1.0003 - 1.73257 I, 1.0003 - 1.73256 I, 1.00029 - 1.73255 I, 
 1.00028 - 1.73254 I, 1.00028 - 1.73253 I, 1.00027 - 1.73252 I, 
 1.00027 - 1.73251 I, 1.00026 - 1.7325 I, 1.00026 - 1.73249 I, 
 1.00025 - 1.73248 I}

Also - not a proof- you see

(Evaluate[(x + 5 Sqrt[4 + x^2])/((9 + 8 x^3)^(1/3)) /. x -> -1000000] // N ) - (1 - I Sqrt[3])

gives

2.50044*10^-12 - 4.33187*10^-12 I

Look also at definition of cubic root.

Cheers,

Marco

PS: Sorry saw Daniel's post too late.

POSTED BY: Marco Thiel
Posted 10 years ago
POSTED BY: Eduen Sarceno

All depends on how that cube root is done. To get surd behavior one requires CubeRoot (or Surd) in Mathematica.

In[154]:= Limit[(x + 5*Sqrt[4 + x^2])/CubeRoot[9 + 8*x^3], 
 x -> -Infinity]

(* Out[154]= -2 *)

In[155]:= Limit[(x + 5*Sqrt[4 + x^2])/(9 + 8*x^3)^(1/3), 
 x -> -Infinity]

(* Out[155]= 1 - I Sqrt[3] *)

As for Wolfram|Alpha, I do not know offhand what heuristics it uses to determine which was intended. But one can force the surd behavior by using cbrt().

lim (x to -infinity ) ((x+5sqr(4+x.b2))/cbrt(9+8x.b3)) enter image description here

POSTED BY: Daniel Lichtblau
Posted 10 years ago

Not really, see you here

POSTED BY: Eduen Sarceno

Hi,

the problem is with your calculation on the board. The solution will be complex because the cubic root in the denominator will go negative as x tends to infinity. The step from your second to the third term and from the third to the fourth appear to be incorrect.

Cheers, Marco

POSTED BY: Marco Thiel
Posted 10 years ago

Not really, see you here

POSTED BY: Eduen Sarceno
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