Find the limit of sequence (trig term)

[izen]

Homework Statement

find the limit of this sequence

1) An = $\frac{\sqrt{n} sin (n! e^{n})}{n+1}$

2) An = $4n^{3} sin\frac{1}{n^{3}}$

Homework Equations

The Attempt at a Solution

(1)
|sin n! e^n)| ≤1

|An| ≤ $\frac{ \sqrt{n} }{n+1}$

|An| ≤ $\frac{ \sqrt{n}/n }{n/n+1/n}$ = 0

lim$_{n\rightarrow \infty}$ An = 0 √

==========================

(2)

$|sin \frac{1}{n^{3}}|$ ≤ 1
An ≤ $4 n^{3}$
lim$_{n\rightarrow \infty}$ An = ∞ X

The second question my answer is wrong
I should get 4. and the answer shows that I have to use L' Hospital's
My question is why I cannot apply the method from the first question to the second question.
It just confused me I am not sure but my solution on the first question anyway but I got the right answer please comment on that too please !

thank you for your help

 

[SammyS]

Homework Statement

find the limit of this sequence

1) An = $\frac{\sqrt{n} sin (n! e^{n})}{n+1}$

2) An = $4n^{3} sin\frac{1}{n^{3}}$

Homework Equations

The Attempt at a Solution

(1)
|sin n! e^n)| ≤1

|An| ≤ $\frac{ \sqrt{n} }{n+1}$

|An| ≤ $\frac{ \sqrt{n}/n }{n/n+1/n}$ = 0

lim$_{n\rightarrow \infty}$ An = 0 √

==========================

(2)

$|sin \frac{1}{n^{3}}|$ ≤ 1
An ≤ $4 n^{3}$
lim$_{n\rightarrow \infty}$ An = ∞ X

The second question my answer is wrong
I should get 4. and the answer shows that I have to use L' Hospital's
My question is why I cannot apply the method from the first question to the second question.
It just confused me I am not sure but my solution on the first question anyway but I got the right answer please comment on that too please !

thank you for your help

As n→∞, 1/n³ → 0 so sin(1/n³) → 0.

So that, 4n³ sin(1/n³) is indeterminate of the form ∞∙0 .

Even if have a sequence such as $\displaystyle 4n^3\sin(n)$ your method will give an incorrect result because sin(n) oscillates from near -1 to near +1 .

 

[Saitama]

For the second question, replace n with 1/h where h->0. Then use L'Hopital's Rule.

 

[izen]

As n→∞, 1/n² → 0 so sin(1/n²) → 0.

So that, 4n³ sin(1/n²) is indeterminate of the form ∞∙0 .

Even if have a sequence such as $\displaystyle 4n^3\sin(n)$ your method will give an incorrect result because sin(n) oscillates from near -1 to near +1 .

What I don't understand is why it works on the first question but the second question. still don't get it. thanks for the comments

 

[izen]

For the second question, replace n with 1/h where h->0. Then use L'Hopital's Rule.

why we can set h->0 I have seen that method but I don't get it why n-> infinity can change to h->0. sorry if I asked such a stupid question. thanks for the comments

 

[haruspex]

As n→∞, 1/n² → 0 so sin(1/n²) → 0.
So that, 4n³ sin(1/n²) is indeterminate of the form ∞∙0 .

The question has 1/n³, not 1/n².
izen, lim _n→∞ n³ sin(1/n³) must be the same as lim _n→∞ n sin(1/n), right? Does that help?

why we can set h->0 I have seen that method but I don't get it why n-> infinity can change to h->0.

It's just a change of variable. Define h = 1/n, then n sin(1/n) = sin(h)/h, and the limit of n sin(1/n) as n→∞ must be the same as lim sin(h)/h as h→0. Having converted it to a problem with a finite limit point you can use l'Hopital's rule.

 

[izen]

I need time to think about your say. Thank you so much haruspex :)

 

[izen]

It's just a change of variable. Define h = 1/n, then n sin(1/n) = sin(h)/h, and the limit of n sin(1/n) as n→∞ must be the same as lim sin(h)/h as h→0. Having converted it to a problem with a finite limit point you can use l'Hopital's rule.

so it means lim n→∞ 1/n = 0 so if we define h=1/n. the limit of 'h' needs to be equal to 1/n by setting lim h→0 h = 0. Am i understand it right ?

 

[SammyS]

What I don't understand is why it works on the first question but the second question. still don't get it. thanks for the comments

Consider a sequence, {A_n}, such that the each individual term is given by,

A_n = B_n∙C_n​

where {B_n} is bounded above and below, but does not converge.

If {C_n} converges to zero, then so does {A_n}.

Otherwise, {A_n} does not converge -- no matter whether {C_n} converges or not.
Your first problem is a case of the above.Your second problem does not nearly fit the above requirements.

Yes, the sequence {sin(1/n³)} is bounded, but more than that, this sequence converges to zero.^* The sequence {4n³} → +∞ as n → ∞ . So, the term by term product of these sequences may or may not converge based upon other totally different criteria.

^* The sequence {1/n³} converges to zero, and sin(0) = 0 .
​

 

[Dansuer]

In your first question you show that your sequence lies beetween two sequence that both goes to zero 0≤An≤0 this implies An → 0
It's important that you show that it lies BEETWEEN two sequences.

In your secondo question you show that your question is less than infinity, that does not make it go to infinity though. An could go 2, which is less than infinity, or -10 which is less than infinity, or -∞ which is less than infinity.

 

[haruspex]

so it means lim n→∞ 1/n = 0 so if we define h=1/n. the limit of 'h' needs to be equal to 1/n by setting lim h→0 h = 0. Am i understand it right ?

Yes.

 

[izen]

i get it now
thank you u guys all