What do you think of my proof for this limit anons?

I'm trying to give a precise proof that the limit is false ,did I succeed in doing so?

seems like you're on the right track but the wording/logic is very awkward

what is N? why is 1+1/n^2

Your proof falls apart for all epsilon greater than 1. For example choose epsilon = 10 then the left hand side equals 4 which is clearly not greater than 10.

To prove a limit is false its best to show that there exists an epsilon greater than zero such that for all natural numbers N, |asub(n)-L|>epsilon if n>N. Try that instead of proving it for all epsilon which most of the time won't work.

>Not proving it's 2

fail/100

?
Lim n -> infinity n^2 (2+(1/n^2) / 1)
= (2+0)/1 = 2, is not equa to 1. Limit is correct.

you have to be 18 to post on this website

fucking this

Same user as above. Your scratch work is good but remember we trying to show that 1+1/n^2>epsilon not less than. So notice that 1+1/n^2 is always greater than 1. So if you choose an epsilon less than 1 then youve sucessfully proven what I recommended to prove above. Try that out.

yup, 1/n^2 would be 1/(1/(epsilon-1)) so in the end you'd get epsilon >= epsilon, OP is bad at math

just did with

idk why he makes the work for himself when it's much easier than that.

are you legitimately retarded?

fuck ya perfect

You are okay but this is kinda pointless.

Just compute the limit and see it does not equal to 1.

The epsilon delta definition of a limit was definitely not made for these kinds of proofs. It is simply supposed to enable the proofs of more complex theorems that then you are supposed to use instead of the definition itself.

Can you stop spamming your shitty brainlet homework here thanks

...

>very fast limit converging at very hihg speed

Top Kek

I don't feel like using LaTex right now..
He's not okay..

How the fuck can 1+(1/n^2) < e
be true for arbitrary epsilon?

If you want to prove that the limit is not 1, then prove that it is 2, which is not equal to 1, which is what the limit is....

That requires a proof of 1=/=2 which can be quite lengthy.

[math]\lim\limits_{n \to +\infty}\frac{2n^{2}+1}{n^{2}} = \lim\limits_{n \to +\infty}\frac{n^{2}\cdot(2+\frac{1}{n^{2}})}{n^{2}} = \lim\limits_{n \to +\infty} (2+\frac{1}{n^2}) = 2[/math]

Thus, it can't be equal to 1.

It really grinds my gears that you factored like that instead of just distributing.

re:the same goes for you.

Why did you set it equal to 2?

FYI, I am this user: The reason why I factored the expression is because I think it's the simplest way to logically explain why said limit is equal to 2, especially considering that the difficulty of the question proposed by OP is relatively easy, thus I have assumed that OP's skills in mathematics are rather undeveloped at this point.

Because [math]\frac{1}{n^{2}}[/math] in this example will tend towards [math]0^{+\infty}[/math] and thus we 'calculate' 2 + 0 = 2. Hopefully this clears it up.

uhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

I am absolutely retarded, I meant [math]0^{+}[/math] of course, not [math]0^{+\infty}[/math].

lmao

Alternatively, you could also do this:

[math]\lim\limits_{n \to +\infty}\frac{2n^{2}+1}{n^{2}} = \lim\limits_{n \to +\infty}\frac{\frac{d}{dn}(2n^{2}+1)}{\frac{d}{dn}(n^{2})} = \frac{4}{2} = 2[/math]

...

Agreeable

Freshman nonsense.

Show that for some n, say n=2, the expression is greater than 1. Then show that for all sufficiently large n, the derivative of the expression is positive. Positive derivative means the function is monotonically increasing, if it's currently bigger than one it thus will always be bigger than one.

Students often try to use more complex machinery than they need to. When you see a problem breathe, consider what's actually going on, and do the simplest thing first.

You are an idiot.

he said nothing wrong and op is a dumb nigger

Yeah I just saw what I did. Pretty silly indeed. I guess that's why one shouldn't rush math exercises

I have no idea what the fuck any of this is.

Every time I come to this board; the thought of suicide follows.

Shouldn't be using the derivative of a fraction instead of what you did?

[math]\frac{2n^2+1}{n^2}=2+\frac{1}{n^2}[/math] which approaches infinity as [math]n \rightarrow \infty [/math]

I did this shit in 10th grade Wtf is your problem?

[math]\frac{2n^2+1}{n^2}=2+\frac{1}{n^2}[/math] which approaches [math]2[/math] as [math]n \rightarrow \infty [/math]

No. He's very correct with his usage of Indeterminate Forms and L’Hospital’s Rule.

L'Hopital's rule, user.

it equals 2. proof: set n to largest length that doesnt cause an overflow

t. engineer

fuck off Bertrand