ATTN: All mathfags

[Anonymous No.6018362]

So i'm reviewing some limits of sequences and shit and came upon this bad boy (pic related).
Now, I'm not the smartest when it comes to remembering these rules but I assume that I am to employ L'Hopitals rule and shit.
When doing this, do I take the product rule of the numerator? I'm fucking confused as shit.

Could be from the lack of sleep tho

>inb4 wolfram

Already tried there but they didn't explain shit cause algorithms and shit

[Anonymous No.6018367]

>>6018362

> pls respond

[Anonymous No.6018374]

>>6018362
Clearly the limit doesn't exist.

[Anonymous No.6018379]

>>6018374
Actually that's wrong. It is indeed zero.

[Anonymous No.6018390]

Throw out the alternating sign term and prove that the rest is positive and decreasing for all n.

[Anonymous No.6018393]

lol sci.

OP: get the 1/n inside the square root:

it become (-1)^n sqrt(1/n^2 + 1/n).
The limit is obviously 0.

[Anonymous No.6018397]

>>6018390
how does one go about doing this?

[Anonymous No.6018409]

>>6018393

the fuck are you talking about

> you can actually put the 1/n in the sqrt term

[Anonymous No.6018411]

>>6018409
if you're asking what the fuck I'm talking about, you have a big problem bro.

<span class="math">\frac{\sqrt{1+n}}{n} = \sqrt{\frac{1+n}{n^2}} = \sqrt{\frac{1}{n^2} + \frac{1}{n}} [/spoiler].

[Anonymous No.6018422]

>>6018411

wow that didn't even cross my mind
> mfw
t-thanks anon

[Anonymous No.6018425]

>>6018422
:D
no pb bro

[Anonymous No.6018480]

>>6018362
I really hope you are not a STEM major.

[Anonymous No.6018561]

solution in 4-space: lim(1→n)=z/((x/y+(2*x*y)+y/x)/(w/z+(2*w*z)+z/w)+(2*(x/y+(2*x*y)+y/x)*(w/z+(2*w*z)+z/w))+y/(x/y+(2*x*y)+(w/z+(2*w*z)+z/w)/x))=e

where (n) is a floating point operator, (e) is the field potential of the universe at the frame of reference(lim(0→e)=1), and (j) is the wavelength of hydrogen-1, defined as 1/n. The series for the rest of the elements follow lim(j→z)=2*j^2+2=n. Now, it's only a floating point operator because when we measure the universe we have to quantify that measurement with reference to (x,y,z,j), and our measurement moves out of one of the 9 imaginary points where (x*y*z*j)=undefined within hyperspace.

we are either one observer throughout all of time, or up to 9 for any quantified point in time, or one quantum observation in 10-d space, or a quantum computer solving the 9-body problem in any-dimensional space.

[Anonymous No.6018569]

>>6018561
frame of reference(lim(n→e)=1)*, oops

[Anonymous No.6018574]

>>6018561
*and hydrogen-1 is 4/e, like the math says.

[Anonymous No.6018620]

Hello!

You could also use the squeeze lemma very effectively here.

Best!

[Anonymous No.6018653]

you could just say 0 because it's completely fucking obvious

[Anonymous No.6018887]

>>6018653
this

[Anonymous No.6019181]

>>6018620

Dis, den L'Hoptail, den limit

[Anonymous No.6019183]

easiest way is to use:
-sqrt(1+n)/n<=f(n)<=sqrt(1+n)/n, then show that both sides go to zero

[Anonymous No.6020936]

>>6018887
Serious this actually takes no math

[Anonymous No.6020939]

Trivial epsilon-n proof. If you can't do this, you probably don't know the definition of sequence convergence and need to re-read the relevant section in your textbook and/or go to office hours.

[Anonymous No.6020944]

>>6020939

Here's the (nonrigorous and sloppy) proof, because I'm bored:

Clearly, sqrt(1+n) <= sqrt(2n) for all n>=1. Sqrt(2n)/n = sqrt2/sqrt(n). The sequence 1/sqrt(n) clearly converges to 0, since for any natural number a we can find another natural number b such that b^2 > a, so the convergence follows from the archimedian property of the real numbers. Since 1/sqrt(n) converges, sqrt(2)/sqrt(n) clearly must also converge, and so must the series in question.

[Anonymous No.6020943]

>>6018620
This please.

[Anonymous No.6020949]

>>6020944

Typo: should be "such that b > a^2."