Limit of sin(1/x) as x approaches infinity: Understanding the Concept

[rambo5330]

Homework Statement

lim (x->inf) sin(1/x)

i have a teacher that seems to think this is equal to 1.. I don't see how this is correct
as x approaches infinity 1/x aproaches zero... sin 0 = 0 right?
or is this the wrong way of thinking?

Homework Equations

The Attempt at a Solution

 

[vela]

You're right. Perhaps your teacher was thinking of

\lim_{x \to 0} \frac{\sin x}{x} = 1

 

[rambo5330]

Thanks very much yes maybe he was thinking of that. but here is the actuall email he sent out correcting himself... if anything it confused me more... does what he say hold true?
"
The application of the squeeze theorem that we did in class to lim(x->inf)sin(1/x)/x, was correct and gives the answer 0, but I was wrong to say that lim(x->inf)sin(1/x) DNE, it is as some students saw 1. What I intended was to consider the limit: lim(x->0) x sin(1/x). Now lim(x->0)sin(1/x) = DNE because as x->0, sin(1/x) oscillates wildly between -1 an +1, so ones really needs the squeeze theorem here!"

 

[seto6]

nop 1/x --->0 so sin(1/x) goes to zero

 

[seto6]

well if u know L'hospital rule you can use it on <br /> \lim_{x \to 0} \frac{\sin x}{x} = 1<br />

so u get <br /> \lim_{x \to 0} \frac{\cos x}{1} = 1<br />