binomial limit (1 Viewer)

tywebb

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I put it into wolframalpha and got
wa.png
 
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tywebb

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This is how I did it without wolframalpha. First use a logarithm to turn the product into a sum so it is easier to deal with. Thereafter it is a matter of simplifying the limit as much as possible.

Method 1 (see more methods in subsequent posts)






 
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synthesisFR

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Ok so its not a 4u question
WHY TF WOULD this person put it in the extension 2 thread i almost was gonna drop 4U looking at this working out
 

tywebb

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The gamma function is just a continuous version of factorial and the digamma and trigamma just come from the derivatives.

The hardest function that appeared was the hyperfactorial but I got rid of it pretty quickly by using l'Hôpital's rule.

That's a nifty rule to use in limits - even for much simpler questions.
 

synthesisFR

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The gamma function is just a continuous version of factorial and the digamma and trigamma just come from the derivatives.

The hardest function that appeared was the hyperfactorial but I got rid of it pretty quickly by using l'Hospital rule.

That's a nifty rule to use - even for much simpler questions.
do u do maths at uni?
 

Pethmin

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idk bruh. It's pretty damn simple 🤓 just limit properties and gamma function is just a continuous version of factorial and the digamma and trigamma just come from the derivatives. hardest function that appeared was the hyperfactorial but I got rid of it pretty quickly by using l'Hospital rule.
That's a nifty rule to use - even for much simpler questions. But realistically speaking if ur doing 4U and u cant even do smn as simple as that bruh 💀 good luck bud. 🤓🤓🤓🤓🤓
 

synthesisFR

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But realistically speaking if ur doing 4U and u cant even do smn as simple as that bruh 💀 good luck bud. 🤓🤓🤓🤓🤓
yeah man...
im considering to drop it tbh, even 3u for that matter. im just not good enough ...
 

tywebb

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There is another way to do it using Stirling's formula

So here is Method 2:







 
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tywebb

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Here is another method using the Stolz–Cesàro theorem. It starts the same as the first method but then goes in a completely different direction, nevertheless ending up at the same answer.

Method 3.



 
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