Calculating limits: limit_{n -> infinity} (n * sin(n!))/(n^2 + 1)

[Qwertyuiop[]]

I have another question: [imath]\lim _{n\to \infty }\left(\frac{n\cdot sin\left(n!\right)}{n^2+1}\right)[/imath]. I used the squeeze theorem to find the limit because [imath]-1\le sin\left(x\right)\le 1[/imath].
[imath]-1\le sin\left(n!\right)\le 1[/imath]
[imath]\lim _{n\to \infty }\left(\frac{-n}{n^2+1}\right)\le \lim \:_{n\to \:\infty \:}\frac{n\cdot sin\left(n!\right)}{n^2+1}\le \lim \:_{n\to \:\infty \:}\left(\frac{n}{n^2+1}\right)[/imath] and the limit of [imath]\left(\frac{-n}{n^2+1}\right)[/imath] and [imath]\left(\frac{n}{n^2+1}\right)[/imath] as n goes to infnity is 0. So [imath]\lim \:_{n\to \:\infty \:}\frac{n\cdot sin\left(n!\right)}{n^2+1} =0[/imath]. But when i check the answer on symbolab , it says "Does not exist". What's wrong with what I did?

 

[BigBeachBanana]

I have another question: [imath]\lim _{n\to \infty }\left(\frac{n\cdot sin\left(n!\right)}{n^2+1}\right)[/imath]. I used the squeeze theorem to find the limit because [imath]-1\le sin\left(x\right)\le 1[/imath].
[imath]-1\le sin\left(n!\right)\le 1[/imath]
[imath]\lim _{n\to \infty }\left(\frac{-n}{n^2+1}\right)\le \lim \:_{n\to \:\infty \:}\frac{n\cdot sin\left(n!\right)}{n^2+1}\le \lim \:_{n\to \:\infty \:}\left(\frac{n}{n^2+1}\right)[/imath] and the limit of [imath]\left(\frac{-n}{n^2+1}\right)[/imath] and [imath]\left(\frac{n}{n^2+1}\right)[/imath] as n goes to infnity is 0. So [imath]\lim \:_{n\to \:\infty \:}\frac{n\cdot sin\left(n!\right)}{n^2+1} =0[/imath]. But when i check the answer on symbolab , it says "Does not exist". What's wrong with what I did?

The answer should be 0. Symbolab isn't the most reliable engine. I prefer WolframAlpha.

 

[topsquark]

I have another question: [imath]\lim _{n\to \infty }\left(\frac{n\cdot sin\left(n!\right)}{n^2+1}\right)[/imath]. I used the squeeze theorem to find the limit because [imath]-1\le sin\left(x\right)\le 1[/imath].
[imath]-1\le sin\left(n!\right)\le 1[/imath]
[imath]\lim _{n\to \infty }\left(\frac{-n}{n^2+1}\right)\le \lim \:_{n\to \:\infty \:}\frac{n\cdot sin\left(n!\right)}{n^2+1}\le \lim \:_{n\to \:\infty \:}\left(\frac{n}{n^2+1}\right)[/imath] and the limit of [imath]\left(\frac{-n}{n^2+1}\right)[/imath] and [imath]\left(\frac{n}{n^2+1}\right)[/imath] as n goes to infnity is 0. So [imath]\lim \:_{n\to \:\infty \:}\frac{n\cdot sin\left(n!\right)}{n^2+1} =0[/imath]. But when i check the answer on symbolab , it says "Does not exist". What's wrong with what I did?

You didn't do anything wrong. Symbolab is probably reacting to the sin(n!), which does not have a limit that exists for [imath]n \to \infty[/imath]. But, as you say, [imath]-1 \leq sin(n!) \leq 1[/imath] and the factor outside of the sine function is [imath]n/(n^2 + 1)[/imath], which has a limit of 0 as [imath]n \to \infty[/imath].

-Dan