# I'm not understanding this limit problem...

• Mar 29th 2009, 09:17 PM
virtuoso735
I'm not understanding this limit problem...
lim of (n)sin(1/n) as n goes to infinity. I thought the limit would be 0 since sin[1/(really big number)] = 0, but the book says the limit is 1. how do you get the limit as 1? Thanks.
• Mar 29th 2009, 09:32 PM
Chris L T521
Quote:

Originally Posted by virtuoso735
lim of (n)sin(1/n) as n goes to infinity. I thought the limit would be 0 since sin[1/(really big number)] = 0, but the book says the limit is 1. how do you get the limit as 1? Thanks.

Let $\displaystyle z=\frac{1}n\implies n=\frac{1}{z}$. Then as $\displaystyle n\to\infty$, $\displaystyle z\to0$. Thus, $\displaystyle \lim_{n\to\infty}n\sin\frac{1}{n}=\lim_{z\to0}\fra c{\sin z}{z}=\dots$
• Mar 29th 2009, 09:52 PM
Quote:

Originally Posted by virtuoso735
lim of (n)sin(1/n) as n goes to infinity. I thought the limit would be 0 since sin[1/(really big number)] = 0, but the book says the limit is 1. how do you get the limit as 1? Thanks.

Dont forget this
Its not just

sin[1/(really big number)]

But its

(really big number) x (sin[1/(really big number)])

So can't just do it that way

However put 1/n =t

So when n-> infinity , t-> 0
Hence it becomes

Lt_{t->0} (sin[t]/t) = 1

For the geometrical proof of this