Question

lim sin(1/n)*n
n->infinity

Answer 1

put 1/n =y
y-> 0
\[\lim_{y\to 0}\frac{\sin y}{y}\]

Answer 2

After that

Answer 3

Should I use L'Hospital rule @ash2326

Answer 4

= 1

Answer 5

Well thanx guys.

Answer 6

it's a standard limit
\[\lim_{x\to 0}\frac{\sin x}{x}=1\]

Answer 7

oh ya.

Answer 8

What if it was
sin(x/n)*n where x is any integer.

Answer 9

As n--->infinity

Answer 10

Let's work with that
\[\lim_{n\to \infty}{\sin x/n}\times n\]
n=1/y
y->0
so we have
\[\lim_{y\to 0}\frac{\sin xy}{y}\]
multiply numerator and denominator by x
\[\lim_{y\to 0}x\frac{\sin xy}{xy}\]
\[x\lim_{y\to 0}\frac{\sin xy}{xy}\]
\[x\times 1\]
\[x\]

Answer 11

Sorry but I think it is x*pi/180

Answer 12

Never mind I think I got it

Answer 13

if x is in degrees, then we'd do that

Answer 14

That depends on what x is.

Answer 15

Thanx @ash2326