Limits

[grave orchid]

I am getting e^(-8) but given answer is e^(-2) pls help.

[lavish goblet]

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[quick cedar]

can you show your work

[naive delta]

i havee the same bedcoverr

[naive delta]

unintelligible

[grave orchid]

Question🥺

[quick cedar]

me too

us BROS us

[naive delta]

sahi se likh le

[grave orchid]

Likha toh hai

[quick cedar]

samjh ni aara

[grave orchid]

Bro u get 1^infinity form so I wrote the standard procedure. That's it

Wait lemme rewrite

[naive delta]

$\lim_{n \to \infty} \cos\left(\ln\left(\frac{n-1}{n+1}\right)\right)^{(n+1)^2}$

[hexed isle]

Coffey

[grave orchid]

Le bhai

[naive delta]

tedha

,rotate

[naive delta]

no idea how you removed the limit

[grave orchid]

😐😐😐😐1^infinity form

[naive delta]

cos(to zero)^(to infty)

[naive delta]

never heard of that before

[grave orchid]

@cosmic quarry can u help

[cosmic quarry]

Let's see.
ln(n - 1) = ln(n) + ln(1 - 1/n)
ln(n + 1) = ln(n) + ln(1 + 1/n)
So:
ln(n - 1) - ln(n + 1) = ln(1 - 1/n) - ln(1 + 1/n)
Next, for n -> ∞ we have:
ln(1 - 1/n) = 1 + 1/n + 1/n^2 + o(1/n^2)
ln(1 + 1/n) = 1 - 1/n + 1/n^2 + o(1/n^2)
So:
ln(1 - 1/n) - ln(1 + 1/n) = 2/n + o(1/n^2), n -> ∞
Then, for x -> 0 we have cos(x) = 1 - x^2/2 + o(x^2). So:
cos(ln(1 - 1/n) - ln(1 + 1/n)) = cos(2/n + o(1/n^2)) = 1 - 2/n^2 + o(1/n^2), n -> ∞
The rest is easy.

[naive delta]

o notation (by darpinger)

[grave orchid]

U won't get the correct answer with this. I just calculated

[cosmic quarry]

You will.

[grave orchid]

U won't get e^-2

[cosmic quarry]

Oh, you saved it! I'm flattered 😁

[cosmic quarry]

You will.

[naive delta]

let's wolfram it

although that's how the limits work

[cosmic quarry]

I can already see that it's e^(-2) from where I stopped.

[cosmic quarry]

Well, you can do it that way too, I suppose. But I like my way here more.

[grave orchid]

What about the power man. U just ignored it. Without power, u get e^(-2) only

There is a n+2 squared power also

[cosmic quarry]

Read the last sentence of what I wrote.

[grave orchid]

But with that u get e^-8 only boss

[cosmic quarry]

No.

[grave orchid]

Pls write it on paper and send bro

I can't understand ur notations

[cosmic quarry]

Well, I can write it, I guess... Won't be much different.

[grave orchid]

Also, what is wrong with my solution

[cosmic quarry]

Oh, I see where you went wrong.
cos(x) - 1 = -2sin(x/2)^2, but what you wrote is basically -2sin(x)^2. Which is why the exponent that you got is 4 times greater.

[grave orchid]

Awww man

Damn

It

!!!!!! I am so stupid!!!!

Thanks a lot bro

But whatever method u used, I didn't get it. In high school, we are only taught one expansion for log(1+x) that is this

Also I didn't understand the o notation

But you were a major help sir thanks a lot!!!

@cosmic quarry

I need help with one more problem. Do u have time?

I need to do this problem without l hospital I did it in the past but I forgot how to approach it without l hospital

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