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?
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Originally Posted by runner07 do we then do L'Hopital's rule? Forget that. What is this limit? $\displaystyle \lim _{n \to \infty } \left( {\frac{\pi }{{2n}}} \right)$ Now answer the question.
the limit of that is 0 isnt it? so we just ignore the sine?
Originally Posted by runner07 so we just ignore the sine? Absolutely not!. The sine function is continuous. Right? What is sin(0)?
the sin(0) is 0....
And that is the sequential limit.
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.....??
$\displaystyle \left( {\frac{{n\pi }}{{4n + 1}}} \right) \to \frac{\pi }{4}\quad \Rightarrow \quad \left[ {\tan \left( {\frac{{n\pi }}{{4n + 1}}} \right)} \right] \to \tan \left( {\frac{\pi }{4}} \right)$
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