Any ideas?

Yeah.. who would have gussed math nerds are the intelligent anti-social type.. unimaginable.

It's upside down.

No? I got nothin.

Guys, I can't use L'hopital's rule, this is exactly why I found this difficult.

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it's 0

applying the rule comes out to:

(10n^9 + 2n)
-----------------
(n^10 + n^2)

...over 1, so that's zero...the formal way

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usually any thread with math in it is just those stupid order of operation argument ones

those arent math threads those are b8 threads and you lost if you dont already have those filtered out

lim < lim ln(2x^10)/x = lim 10ln(x)/x = 0

I can use wolfram|alpha too, I'm asking about the solution.

Why can't atheists define atheism?

Bruh, its just hierarchy of functions. Factorials piss on exponentials, but these butt fuck polynomials, and polynomials bitch slap logs. So when you see ratios the one on top of the pyramid always grows infinetly faster.

there are several already in the thread, don't be a dick about it when Cred Forumstards are actually helpful

>cursive

what is this, 4th grade?

2nd picture I showed you explains why.

i did it in my head m'dude, ease up on the salt

That didn't derive correctly.

I said I couldn't use L'Hopital's

Do you even the graph of natural logarithm?

That's what I was thinking, but I am too lazy to see what it turns out to be. Use L'Hopital's rule to get the nautural log into a fraction and then you should be able to divide it out into a nice working limit.

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Actually, how the brown did that system even apply L'Hopital?

Where, when, why not?

Then you're going to have to use the limit definition. Are you familiar with it?

Or he could try a qualitative approach by graphing the function and entering a really large number as an n input.

n^10>>n^2;
nSo your limit is equivalent to ln(n^10)/n=10ln(n)/n->0. Proof by l'hospital rule:ln(n)'/n'=1/n->0.
Swag yolo

We're not supposed to know what is a derivative, only the definition of limit (not even that of a function, we're working with sequences) and some standard limits like lim (1+1/n)^n = e

Ofc, and this is the only thing that we should be aware of when solving limits, according to our calculus teacher

Well before we try to help anymore you should provide the full context of the problem - what theorems are you allowed to use? Or do you have to use the definition?

So effectively you just need to prove that e^n>>n.

So you must show that for all epsilon > 0 there exists m in N such that, for all n in N, n > m implies | ln(n^10)/n -

We can use the definition, limit properties, limits of common sequences and theorems about sequences not involving derivatives, integrals etc. like Stolz-Cesaro theorem or Weierstrass's theorems

ever heard of a graph?

I need the exact solution fagtron

As n->inf the denominator approaches inf. The numerator approaches ln(inf) which is also infinity, meaning as a whole the function's limit is 1. inf/inf=1

So n/2n->1 as n->oo? Fucking loser.

Don't be rude on my behalf you faggot. He's not OP I can prove

If I will it to it eventually does.

so edgy.

I've had enough of your attitude op. I have the solution you desire but won't post until you apologize to that user.

man i remember middleschool

Why are you on an 18+ board

I see 1/(n-2) something you don't go over in baby's calc1 I also see log rules.

If I'm not mistaken it's log. You can do a quick proff too