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How do you solve this limit?

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Thread replies: 18
Thread images: 4

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LIMITLESSS.jpg
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I know it diverges, but I don't find a way to prove it. I've tried using ln on both sides but i get to a plece where I get 0*infinity and There's nothing left to wotk with.
Any idea?
>>
Try taking the derivative of it.
>>
>>18052174
Use Stirling approximation
>>
L'Hopital rule i believe should work
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>>18052174

You should be able to solve it from here, friend!
>>
>>18052211
Are you sure you can take the limit out of the logarithm, m8?
I suggest being careful with such things.
>>
>>18052226
Yes, log is a monotonically increasing function. It's the same reason why you can take the limit out of e.
>>
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lel m8
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>>18052371
fuck i am retarded and missed factorial, w8
>>
Note that (n/2)^(n/2) < n!

So taking the nth root of both sides gives:

Sqrt (n/2) < n!^(1/n)

Take the limit of both sides as n goes to inf and you see that the left side goes to inf. Therefore the right side must be inf as well.
>>
>>18052195
That's a way to do it, but isn't there something more simple?
>>
>>18052211
We can take n! as n(n-1)... And then have ln(n)+ln(n-1)... But there's nothing left to do because we can't distribute the limit if there's something wrong and each term gives infinity over infinity
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>>18053508
Yeah, I just showed you here:
>>18052616
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>>18053967
Yeah, yours seems like the most easy approach but how do you know that the inequality that you proposed holds true? It isn't that evident
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>>18054636
n!=n*(n-1)*(n-2)*....*(n/2)*...*2*1
>n*(n-1)*(n-2)*....*(n/2)
>(n/2)*(n/2)*....*(n/2) = (n/2)^(n/2)
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>>18054893
Shit those are supposed to be greater than signs, not greentext
>>
>>
Wait, isn't the nth root of n! just (n-1)!

In that proving that it diverges is super ez
Thread posts: 18
Thread images: 4


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