What do I do

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[lunar valve]

it has to be <1

|x|<1 correct

so solve that

x is the expression you have here

oh and take the limit as n->infinity

[silver gust]

Yeah idk how to solve this

I’m just stuck on that

[lunar valve]

oh

[lunar valve]

show your working

[silver gust]

I did

In the picture is all I got

[lunar valve]

okay

so $\lim_{n\to\infty}\bigg\lvert\frac{2(x+1)\sqrt{n+1}}{\sqrt{n+2}}\bigg\rvert<1$

[round atlas]

dark matter

[lunar valve]

we can take the 2(x+1) our

out

to get $2(x+1)\bigg\lvert\lim_{n\to\infty}\frac{\sqrt{n+1}}{\sqrt{n+2}}\bigg\rvert<1$

[round atlas]

dark matter

[lunar valve]

so what is that limit?

[silver gust]

Bruh I didn’t know u could do that

[silver gust]

Idk how u solve that

Inf /inf?

[lunar valve]

uh

divide by sqrt(n) on the numerator and denominator

then compute the limit

[silver gust]

Yeah idk how to

[lunar valve]

you can since the subject of of the limit is n

[silver gust]

Sqrt(n+1)/sqrtn

[lunar valve]

uh the 2(x+1) should be in the absolute value

[silver gust]

Idk this

[lunar valve]

okay what about this limit

lim as n->infinity of (n+1)/(n+2)

[silver gust]

Times 1/n?

[lunar valve]

yeah

then compute it

or use the asymptote rule

[silver gust]

So this is =1

[silver gust]

But this has Sqrt

[lunar valve]

there is a nice limit property

$\lim_{x\to a}\sqrt{f(x)}=\sqrt{\lim_{x\to a}f(x)}$

[round atlas]

dark matter

[lunar valve]

for sufficient f(x)

in this case it suffices

so what is the limit inside

[silver gust]

Idk

[silver gust]

Howd u do this tho

[lunar valve]

$\lim_{n\to\infty}\frac{\sqrt{1+\frac{1}{n}}}{\sqrt{1+\frac{2}{n}}}$

[round atlas]

dark matter

[silver gust]

u can times by 1/n even if its inside the sqrt?

[burnt tulip]

yes, because n is positive

sqrt(n+1)/sqrt(n+2) = sqrt(n (1+1/n)) / sqrt(n (1+2/n)) = (sqrt(n) · sqrt(1+1/n)) / (sqnt(n) · sqrt(1+2/n)) = sqrt(1+1/n) / sqrt(1+2/n)

[silver gust]

So ur saying if it’s something like (n+1)^2 I can just times the n and 1 inside by 1/n

[burnt tulip]

yes, if it is (n+1)^2, you say that it is (n+1)^2 = (n (1+1/n))^2 = n^2 · (1+1/n)^2

basically, it works with any power c. The property you are using is that for all positive x and y, then you have (xy)^c = x^c y^c

then (n+k) = n · (1+k/n), and then you apply this property to this product

[silver gust]

So if I times 1/n to (1+n)^2 it becomes (1/n + 1)^2? Because sqrt(n+1) is basically (n+1)^1/2

[burnt tulip]

no, you first factor the inner expression by n, and then you get the n^c out

so for (1+n)^2, you would need to time 1/n^2 to (1+n)^2 to get (1+1/n)^2

the rule is basically : if you multiply a by b^c, you first write a = (a^(1/c))^c, then you have a · b^c = (a^(1/c))^c · b^c = (a^(1/c) · b)^c

[lunar valve]

@silver gust take your time to process it

[silver gust]

What about this

[silver gust]

Ok so I don’t actually times the equation by 1/n ur just factoring it out

[burnt tulip]

exactly

[burnt tulip]

you compute this integral using integration by parts, but it might be better to open another thread to discuss it

(I'm leaving now btw)

[lunar valve]

yeah do what petaire said integration by parts

preferrably rewrite $\frac{1}{e^{2x}}$ as $e^{-2x}$ to make it clearer

[round atlas]

dark matter

[lunar valve]

so you;re only integrating $\int (1+x)e^{-2x}dx$ and it is pretty clear what to choose to differentiate and integrate

[round atlas]

dark matter

[silver gust]

Omg bruh I can never figure these out without help

Like damn

No wonder I failed my class

I make u=(1+x) right?

[lunar valve]

yes

[silver gust]

+close

[gloomy rune]

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[gloomy rune]

@silver gust

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