/sci/ - Science & Math

>>10715630
>I'm defending my high school graduation thesis

>>10715634
Eastern European school system

>>10715630
1

>>10715675
AAAAAAAAAAAAAAAAAAAAAAAA NOOOO JOKESSSSSS PLEASS

>>10715675
I got that too but wolframalpha disagrees.

https://www.wolframalpha.com/input/?i=lim+n-%3Einf+((2n%2B2)%2F(2n-1))%5En

>>10715630

>>10715630
>literal state of /sci/.

>high school graduation thesis

Easy. Just create an analytical essay on how incorporating more minority students into the curriculum will somehow improve the educational satisfaction for students of all ethnicities.

>>10715699
Limit of x as x goes to infinity. What does that have to do with the equation?

>>10715630
>>10715699
Whoops, it should be n that is approaching infinity, not x

>>10715713
*limit of f(x) as x goes to infinity

>>10715714
ur n looks like m

Thanks guys, and yeah the x was a typo
But I don't understand the last line, how did you get e^3/2

I mean how did you get e^3/2 from lim n to infnty etc

>>10715734
>>10715733

That is the main formula
but how the fuck from 3n/2n-1 i get 3/2
shouldn't it be 3/2-3n or i'm missing sth

goddamnit dunno what i'm missing AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

>>10715630
The expression in your image is trivial to solve for.
[math]
\displaystyle \lim_{x\to\infty} \left(\frac{2n+2}{2n-1}\right)^n = \left(\frac{2n+2}{2n-1}\right)^n
[/math]
However, if we were meant to find the limit as [math]n\to\infty[/math] instead:
[math]
\begin{align}
\lim_{n\to\infty} \left(\frac{2n+2}{2n-1}\right)^n &= \exp{\left(\log{\left(\lim_{n\to\infty} \left(\frac{2n+2}{2n-1}\right)^n\right)}\right)} \\
&= \exp{\left(\lim_{n\to\infty} \log{\left(\left(\frac{2n+2}{2n-1}\right)^n\right)}\right)} \\
&= \exp{\left(\lim_{n\to\infty} n\log{\left(\frac{2n+2}{2n-1}\right)}\right)} \\
&= \exp{\left(\lim_{n\to\infty} \frac{\log{\left(\frac{2n+2}{2n-1}\right)}}{1/n}\right)} \\
&= \exp{\left(\lim_{n\to\infty} \frac{\frac{d}{dn}\left(\log{\left(\frac{2n+2}{2n-1}\right)}\right)}{\frac{d}{dn}\left(1/n\right)}\right)} \\
&= \exp{\left(\lim_{n\to\infty} \frac{-\frac{3}{2n^2+n-1}}{-1/n^2}\right)} \\
&= \exp{\left(\lim_{n\to\infty} \frac{3}{2n^2+n-1}\left(\frac{1}{1/n^2}\right)\right)} \\
&= \exp{\left(\lim_{n\to\infty} \frac{3}{2+\frac{1}{n}-\frac{1}{n^2}}\right)} \\
&\boxed{= \exp{\left(\frac{3}{2}\right)} = e^{3/2}}
\end{align}
[/math]
I can explain each and every individual line thoroughly if you like. Since I myself just graduated high school a week ago, I'd think you could understand it as well, but I won't make any assumptions or judgments. We're all here to learn. (Well, you and I are. Others are here to shitpost.)
The main crux of the thing is that [math]1^{\infty}[/math] is an indeterminate form; there are uncountably many limits that "look like" that and have different values, so if you plug whatever the variable is approaching into the limit and get [math]1^{\infty}[/math], that doesn't tell you what the value of the limit actually is. You have to apply special techniques to solve for the limit, which is exactly what I did. Again, I can explain it all. Just ask.

>>10715745

>>10715760
>>10715761
Okay, I finally understand now, thanks a lot.
In a few hours I should defend my "work" so I was kinda getting scared, it would be real embarassing if I got a wrong answer
Seriosly you guys saved my ass

>>10715773
this isn't even a thesis. What kind of bullshit high school are you even attending. God damn stupid bait post.
>but zis is 4chan. are you a newfaggerino? this was just a bait, i'm not stupid hahahahaha.
Fuck off.

This is a part of it, I'm going to a gymnasium in Belgrade, this a expression from about 30 others

Ofcourse it isn't a serious academical work, this should be my first encounter with a scientific work, so I can get to the Academy
It is a highly flawed system, but a system nevertheless

>>10715773
No problem.
If you're still not totally clear on any concepts, again, just let me know.
In particular, if you don't quite get the whole indeterminate form thing: essentially, if you have a limit that takes on an indeterminate form, that isn't enough information to tell you the actual value of the limit.
The seven indeterminate forms we consider are: [math]\displaystyle \boxed{\frac{0}{0},\ \frac{\infty}{\infty},\ 0 \cdot \infty,\ 1^{\infty},\ \infty - \infty,\ 0^0,\ \text{and}\ infty^0.}[/math]
For example:
[math]
\begin{align}
\lim_{x\to0} \frac{x}{x^3} &= \infty \\
\lim_{x\to0} \frac{x}{x} &= 1 \\
\lim_{x\to0} \frac{x^2}{x} &= 0
\end{align}
[/math]
These all take on the form [math]0/0[/math], but they all have different values. Just knowing that they take on [math]0/0[/math] isn't enough to ascertain their value, since it's indeterminate.

In our specific case before, we had [math]1^{\infty}[/math]. Let's see:
[math]
\begin{align}
\lim_{x\to\infty} 1^x &= 1 \\
\lim_{x\to1} x^{1/{\log_10{(x)}}} &= 10 \\
\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x &= e
\end{align}
[/math]
So you'll see that [math]1^{\infty}[/math] is also an indeterminate form, and knowing that a limit takes on it isn't enough to know the limit's value.

So what do we do with really complicated limits that take on indeterminate forms? Well, the most common technique to take care of them is L'Hospital's rule.
L'Hospital's rule states: [math]\displaystyle \boxed{\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)}\ \text{if}\ \lim{x\to c} \frac{f(x)}{g(x)} = \frac{0}{0}\ \text{or}\ \pm\frac{\infty}{\infty}}[/math].
This is actually even more helpful than it looks since, with the help of exponentials and logarithms, any indeterminate form can be transformed into either [math]0/0[/math] or [math]\infty/{\infty}[/math]. You'll notice I used it in my previous post.

>>10715773
No problem.
If you're still not totally clear on any concepts, again, just let me know.
In particular, if you don't quite get the whole indeterminate form thing: essentially, if you have a limit that takes on an indeterminate form, that isn't enough information to tell you the actual value of the limit.
The seven indeterminate forms we consider are: [math]\displaystyle \boxed{\frac{0}{0},\ \frac{\infty}{\infty},\ 0 \cdot \infty,\ 1^{\infty},\ \infty - \infty,\ 0^0,\ \text{and}\ infty^0.}[/math]
For example:
[math]
\begin{align}
\lim_{x\to0} \frac{x}{x^3} &= \infty \\
\lim_{x\to0} \frac{x}{x} &= 1 \\
\lim_{x\to0} \frac{x^2}{x} &= 0
\end{align}
[/math]
These all take on the form [math]0/0[/math], but they all have different values. Just knowing that they take on [math]0/0[/math] isn't enough to ascertain their value, since it's indeterminate.

In our specific case before, we had [math]1^{\infty}[/math]. Let's see:
[math]
\begin{align}
\lim_{x\to\infty} 1^x &= 1 \\
\lim_{x\to1} x^{1/{\log_7{(x)}}} &= 7 \\
\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x &= e
\end{align}
[/math]
So you'll see that [math]1^{\infty}[/math] is also an indeterminate form, and knowing that a limit takes on it isn't enough to know the limit's value.

So what do we do with really complicated limits that take on indeterminate forms? Well, the most common technique to take care of them is L'Hospital's rule.
L'Hospital's rule states: [math]\displaystyle \boxed{\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)}\ \text{if}\ \lim_{x\to c} \frac{f(x)}{g(x)} = \frac{0}{0}\ \text{or}\ \pm\frac{\infty}{\infty}}[/math].
This is actually even more helpful than it looks since, with the help of exponentials and logarithms, any indeterminate form can be transformed into either [math]0/0[/math] or [math]\infty/{\infty}[/math]. You'll notice I used it in my previous post.

>>10715773
ur so dumb wtf

>>10715781
He's trolling

>>10715714
why do your ones look retarded?
also I did it too but used l'Hospital's rule
but I see what you did using the formal definition of e there, very clever manipulation.