If x is approaching 0, 1/x is approaching infinity or -infinity, depending if x is positive or negative. What does sin do if the value goes to infinity? -infinity?Homework Statement
Find the limit of sin(1/x) as x -> 0, I don't really understand why there isn't a limit, wouldn't it be +/- 1? or it doesn't exist because there are two values? I just need a more thorough explanation of the answer really.
You're looking at this way too hard. Let's look at a similar example. If you have sin x as x-> infinity, we get oscillation between -1 and 1 and this occurs forever so the limit does not exist. So now looking at sin 1/x as x-> 0, 1/x goes can go infinity or negative infinity, so...no it approaches 0... I see now that the reciprocal makes it larger while the values approaching infinity make it signifigantly smaller. But wait, aren't real numbers paired with circular functions interpreted as radians if pi isn't present? Wouldn't this just mean the value of x represents a reference arc of 360 degrees or 2pi? How does that value go off unto infinity when it's domain is locked in at [-1,1] such that it's amplitude is equal to one. I'm sorry if how this is in some way phrased unclearly.
If you can, look at the graph of tangent. From -pi to pi, the y values range from -infinity to +infinity. From pi to 3pi, the y values range from -infinity to +infinity, ...I see, the infinite behavior arises from the fact that these are waves. Now let me ask, how would this change if the function was tangent or cotangent? the domain isn't continuous. Where there are asymptotes at every n or n/2 value of pi. How could the limit be interpreted if the range was infinity?