Question

Help with Calculus please!
Find: lim as n approaches infinity of nsin(pi/n)

Answer 1

i think you can use squeeze thm

Answer 2

squeeze?

Answer 3

also know as pinch
or even sandwich theorem

Answer 4

or maybe a series expansion that would be fun :)

Answer 5

lim n->∞ sin(PI/n) / (1/n)
0/0 form : Use L'Hopital's Rule
lim n->∞ (-PI /n^2)cos(PI/n) /(-1/n^2)
lim n->∞ PI cos(PI/n)
as n->∞ , cos(PI/n) -> cos(0)=1
so the limits is 'PI'

Answer 6

=lim n->infinity sin(PI/n) / (1/n)
0/0 form : Use L'Hopital's Rule
=lim n->infinity (-PI /n^2)cos(PI/n) /(-1/n^2)
= lim n->infinity PI cos(PI/n)
as n->infinity , cos(PI/n) -> cos(0)=1
so the limits is 'PI'

Answer 7

thanks gorv
but is there a way to solve it without bringing in cos?

Answer 8

substitution would work
recall
\[\lim_{u \rightarrow 0}\frac{\sin(u)}{u}=1\]
\[\lim_{x \rightarrow \infty} \pi \frac{\sin(\frac{\pi}{x})}{\frac{\pi}{x}}\]
let u=pi/x
so if x goes to infty, then u goes to 0

Answer 9

L'Hopital's Rule use derivative thatz why cos comes here derivative of sin

Answer 10

thanks frecles
gorv, so if cos(0)=1, why is the limit pi?

Answer 11

i think he meant to write pi*cos(pi/n)
the inside goes to 0
so you have pi*cos(0)
pi*1=pi

Answer 12

you know because he had (-pi/x^2)/(-1/x^2)
he canceled out the -1/x^2's
but he still had a pi that he probably forgot to type in but it is meant to be there in that one line

Answer 13

or actually he didn't really make an error in typing in anything he said

Answer 14

you know because he didn't use = for each line

Answer 15

\[\lim_{n \rightarrow \infty}\frac{\sin(\frac{\pi}{n})}{\frac{1}{n}} =\lim_{n \rightarrow \infty}\frac{\frac{-\pi}{n^2} \cos(\frac{\pi}{n})}{\frac{-1}{n^2}}\]

Answer 16

see how the -1/n^2 's cancel?

Answer 17

no T_T

Answer 18

\[\lim_{n \rightarrow \infty} \pi \frac{\frac{-1}{n^2} \cos(\frac{\pi}{n})}{\frac{-1}{n^2}}\]

Answer 19

how about now?

Answer 20

yes

Answer 21

so we see how we have this:
\[\lim_{x \rightarrow \infty} \pi \cos(\frac{\pi}{n})\]

Answer 22

yes but I thought it was ((-pi/n^2)cos(pi/n))/(-1/n^2)

Answer 23

the -1/n^2's canceled

Answer 24

but there's only one (-1/n^2). how does (-pi/n^2) and (-1/n^2) cancel?

Answer 25

-pi/n^2 is equal to pi times -1/n^2

Answer 26

\[\lim_{n \rightarrow \infty}\frac{\sin(\frac{\pi}{n})}{\frac{1}{n}} =\lim_{n \rightarrow \infty}\frac{\frac{-\pi}{n^2} \cos(\frac{\pi}{n})}{\frac{-1}{n^2}} \\=\lim_{n \rightarrow \infty} \frac{\pi \cdot \frac{-1}{n^2} \cos(\frac{\pi}{n})}{\frac{-1}{n^2}}=\lim_{n \rightarrow \infty}\frac{\pi \cdot \cancel{\frac{-1}{n^2}} \cos(\frac{\pi}{n})}{\cancel{\frac{-1}{n^2}}}\]

Answer 27

omg thank you!!

Answer 28

np