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The limit of a trigonometric function.

  • Thread starter icesalmon
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  • #1
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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.
 

Answers and Replies

  • #2
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Essentially, yes the limit doesn't exist because there are two values. More specifically, the sine function is oscillating as you extend towards infinity; not just between its two extreme values (1 and -1), but everything in between as well.
 
  • #3
gb7nash
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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.
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?
 
  • #4
244
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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.
 
  • #5
gb7nash
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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.
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...
 
  • #6
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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 npi or npi/2 value. How could the limit be interpreted if the range was infinity?
 
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  • #7
gb7nash
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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?
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, ...

Also notice that (every odd multiple of pi)/2 is a vertical asymptote

So if x approaches infinity, there's no way we're going to get an actual limit for tangent.
 
  • #8
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yeah it makes more sense now, interpreting the limits of circular functions, thank you for your time!
 

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