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

1. ## 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.

2. 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, $\displaystyle \lim_{n\to\infty}n\sin\frac{1}{n}=\lim_{z\to0}\fra c{\sin z}{z}=\dots$

3. 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