1. ## Sequences...

it says...

State whether the sequence converges and if it does, find the limit.

sin[(pi)/(2n)]

so do we do this..?

lim (sin[(pi)/(2n)])
n->oo

but im stuck there. do we then do L'Hopital's rule?

2. Originally Posted by runner07
do we then do L'Hopital's rule?
Forget that.
What is this limit?
$\displaystyle \lim _{n \to \infty } \left( {\frac{\pi }{{2n}}} \right)$

3. the limit of that is 0 isnt it? so we just ignore the sine?

4. Originally Posted by runner07
so we just ignore the sine?
Absolutely not!.
The sine function is continuous. Right?
What is sin(0)?

5. the sin(0) is 0....

6. And that is the sequential limit.

7. oh. does that work for all trig? like if its something like
tan{[(n)(pi)] / (4n+1)}
i would take the limit of pi(n)/ 4n+1... and i'd do L'Hopital's rule.... so it'd be lim pi/4.....??

8. $\displaystyle \left( {\frac{{n\pi }}{{4n + 1}}} \right) \to \frac{\pi }{4}\quad \Rightarrow \quad \left[ {\tan \left( {\frac{{n\pi }}{{4n + 1}}} \right)} \right] \to \tan \left( {\frac{\pi }{4}} \right)$