What is the meaning of this notation?

[WMDhamnekar]

In hp 50g emulator, I performed this computation $\lim\limits_{x\to\infty} [2^{2x} \times (\frac13)^x + (\frac12)^{2x}\times (\frac23)^x ]$

What is the meaning of +:0 ?

 

[BvU]

Did you try to estimate the answer ?
Another thing you can do: look at the terms separately

$\$

 

[malawi_glenn]

Try to read the manual https://literature.hpcalc.org/official/hp50g-um-en.pdf

 

[WMDhamnekar]

Try to read the manual https://literature.hpcalc.org/official/hp50g-um-en.pdf

Would you tell me on which page(page number), I shall get the information about +:0 ?

 

[malawi_glenn]

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 $+ \infty$

 

[jedishrfu]

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 The "hp calculator" results superseded everything else.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.

 

[Mark44]

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.

 

[jedishrfu]

I found it in the HP 50G User Guide in the big version with 887 pgs

https://www.hpcalc.org/details/6512

 

[WMDhamnekar]

I found it in the HP 50G User Guide in the big version with 887 pgs

https://www.hpcalc.org/details/6512

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That means, in our context, the limit is slightly higher than 0 when $x \to \infty$ which is wrong. hp 50g emulator gave wrong answer because its stacks are overflowed when maxR > +9E499 or minR < -9E499.

 

[Tom.G]

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.