Question

Limit of the following function as n approaches zero:

Answer 1

\[nsin(\frac{ 1 }{ n })=?\]

Answer 2

n approaches infinity''' can somebody help me understand this problem?

Answer 3

big hint :)
\[-1 \le \sin(\frac{1}{n} )\le 1\]

Answer 4

oh are you doing two different problems?

Answer 5

I see n->0 in the title
and n->infty in your last post

Answer 6

I meant to say infinity

Answer 7

\[\lim_{n \rightarrow \infty}\frac{\sin(\frac{1}{n})}{\frac{1}{n}}\]
if n->infty then 1/n->?

Answer 8

\[\lim_{u \rightarrow 0}\frac{\sin(u)}{u}=?\]

Answer 9

zero, the fraction becomes infinitely small

Answer 10

so as u->0 then sin(u)/u->?

Answer 11

there is a famous limit proved in many calculus books

Answer 12

sin(u)/(u) would yield 0 over 0

Answer 13

well i mean sounds like you want to do l'hospital now
instead of just remembering by sandwish theorem we have
sin(u)/u->1 as u->0

Answer 14

so we have the following inequality
\[\text{ area of the smaller triangle } \le \text{ area of sector } \le \text{ area of bigger triangle }\]
\[\frac{1}{2} \sin(\theta) \cos(\theta) \le \frac{ \theta}{2} \le \frac{1}{2} \tan(\theta) \\ \sin(\theta) \cos(\theta) \le \theta \le \frac{\sin(\theta)}{\cos(\theta)} \\ \text{ assume } \sin(\theta)>0 \\ \text{ dividing both sides by } \sin(\theta) \\ \cos(\theta) \le \frac{ \theta}{ \sin(\theta)} \le \frac{1}{\cos(\theta)} \\ \lim_{\theta \rightarrow 0^+}\frac{ \sin(\theta)}{\theta}=1\]
we can show in a similar way from the left we also have 1