Requesting to solve this equation. There's an image made

>>126190
type that shit into wolfram alpha
here's a simpler example
http://www.wolframalpha.com/input/?i=lim+x+to+infinity+1%2Fx

>>126195
Math and calculations aren't my thing. If the answer is 0, then it's probably incorrect because I still can't access the file.

>>126197
I wasn't suggesting that the answer to the shit you posted was zero.
Rather the limit of 1/x as x goes to infinity (a simpler example of a limit) is zero.
I wanted you to use the syntax of the example to get wolfram to do the work for you.

see this is closer to the expression you posted
https://www.wolframalpha.com/input/?i=lim+h+to+infinity+%281%2B+%28243%2Fh%29%29^h

>>126197
>here's a simpler example

Means he didn't put in the exactly what you asked for read before you write.

http://www.wolframalpha.com/input/?i=lim+x+to+infinity+%281%2B243%2Fx%2B2520%2F%28x^2%29%2B3216%2F%28x^3%29%29^%28x%2F3%29

I'd say it's equal to 1 but I'm not really sure about this

e^81 ?

>>126200
>>126201
>>126203
>>126204
I entered 1, nothing. I entered e^81, nothing. I entered 1.5060973e+35, nothing. Image is still unlocked.

>>126206
what is that below the expression?
hiragana?
maybe you need to incorporate that as well

>>126207
の上ワケタがパスワード

I think it means the above equation is the password. Not sure about ワケタ (Waketa). It means to divide into, to split into, to separate, to sort out, to classify.

File: wrongmaybe.png (32KB, 676x446px)

32KB, 676x446px

This is probably wrong, but this is all I got from chucking it into wolfram.

>>126213
you used both x and h, that doesn't jibe with op's pic

>>126213
oh shit, i mixed up x and h. It comes out as e^81 instead of 1.

>>126211
7ケタ not ワケタ

>>126216
Author says "input seven upper digits,please."

OP here. The answer is 1.5060973e+35 but the author stated the answer is 7 digits only, so "1506097".

Thanks for the help, everyone.

First it's easy to find a lower bound:
lim (1 + (243/h) + (2520)/h^2 + (3216/h^3))^(h/3)
>= lim (1 + (243/h))^(h/3)
= e^81

Now for every epsilon>0 there is a h_0 such that for all h > h_0
(2520/h) + (3216/h^2) < 3*epsilon
Therefore
lim (1 + (243/h) + (2520)/h^2 + (3216/h^3))^(h/3)
<= lim (1 + (243/h) + (3*epsilon/h))^(h/3)
= e^(81 + epsilon)
Since this is true for every epsilon>0 we must have
lim (1 + (243/h) + (2520)/h^2 + (3216/h^3))^(h/3) <= e^81

Together with the lower bound this gives
lim (1 + (243/h) + (2520)/h^2 + (3216/h^3))^(h/3) = e^81

>>126206
Have you considered that the image is just a way for them to remember their password and that the password is not actually the solution to the problem?

My password recovery questions are rarely answered using legitimate information about me, especially easily obtainable information.

File: 1442284582894.jpg (94KB, 775x719px) Image search: [Google]

94KB, 775x719px

>>126218
enter 2432520

those are the 7 upper digits