- #1

izen

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## Homework Statement

find the limit of this sequence

1) An = [itex]\frac{\sqrt{n} sin (n! e^{n})}{n+1}[/itex]

2) An = [itex] 4n^{3} sin\frac{1}{n^{3}} [/itex]

## Homework Equations

## The Attempt at a Solution

(1)

|sin n! e^n)| ≤1

|An| ≤ [itex]\frac{ \sqrt{n} }{n+1}[/itex]

|An| ≤ [itex]\frac{ \sqrt{n}/n }{n/n+1/n}[/itex] = 0

lim[itex] _{n\rightarrow \infty}[/itex] An = 0 √

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

(2)

[itex]|sin \frac{1}{n^{3}}| [/itex] ≤ 1

An ≤ [itex] 4 n^{3} [/itex]

lim[itex] _{n\rightarrow \infty}[/itex] 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