What is the meaning of this notation?

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In summary, The limit of your expression is +infinity. hp 50g emulator gave wrong answer because its stacks are overflowed when maxR > +9E499 or minR < -9E499.
  • #1
WMDhamnekar
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In hp 50g emulator, I performed this computation ## \lim\limits_{x\to\infty} [2^{2x} \times (\frac13)^x + (\frac12)^{2x}\times (\frac23)^x ]##
1678520543917.png
What is the meaning of +:0 ?
 
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  • #2
Did you try to estimate the answer ?
Another thing you can do: look at the terms separately

##\ ##
 
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  • #5
WMDhamnekar said:
Would you tell me on which page(page number), I shall get the information about +:0 ?
Can you make an effort yourself? I just suggested that the manual would contain the information you were looking for.

Have you computed limits on this calculator before? Have you tried a limit that you know what it should be?

Clearly the limit of your expression is ##+ \infty##
 
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  • #6
I think the expression means you are approaching the limit from the right ( +:0 ) vs ( -:0 ) for approaching from the left.

As an aside, I searched for this +:0 using Google search (hp calculator +:0), and nothing came up. The "hp calculator" results superseded everything else.

Looking back in the manual provided by @malawi_glenn on page 11.2, I did find a reference to using +0 and -0 but not +:0 or -:0.

As a student or programmer, I wouldn't have guessed that usage from +0 or -0 in the manual.
 
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  • #7
jedishrfu said:
I think the expression means you are approaching the limit from the right ( +:0 ) vs ( -:0 ) for approaching from the left.
I don't think that makes any sense, as the limit is as x approaches infinity. And per @malawi_glenn, the limit itself is infinity.
 
  • #10
WMDhamnekar said:
gave wrong answer because its stacks are overflowed when maxR > +9E499 or minR < -9E499.
In our finite world (Universe?) I doubt it makes much difference. 🤪
 
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