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Does lim x →0 sin 1/x exist? also .

  • Thread starter hermy
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  • #1
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Does lim x →0 sin 1/x exist? also.....

Homework Statement



does lim x →0 sin(1/x) exist?

The Attempt at a Solution



I am very new to calculus. From what i have understood till now, limit of a function at a point exists only if the left and right hand limits are equal. What would the graph of sin(1/x) look like? Would the left and right hand limits exist? Do they exist in all cases, for all functions?
 

Answers and Replies

  • #2
matt grime
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What would the graph of sin(1/x) look like?
try drawing it and find out.

Would the left and right hand limits exist? Do they exist in all cases, for all functions?
It would be a bit dull if left and right limits always exist. Remember that limits are unique. So, you need to think about sequences of real numbers x_n tending to zero.

Alternatively we can think about

lim sin(y)

as y tends to infinity.
 
  • #3
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Btw, the existence of left and right limits of a function at a certain point is necessary but not sufficient for a limit to exist at that point. Left and right limits should also be equal for a limit to exist at the same point.
 
  • #4
HallsofIvy
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In particular, think about the sequences [itex]x_n= 1/(n\pi)[/itex], [itex]x_n= 2/((4n+1)\pi)[/itex], and [itex]x_n= 2/((4n-1)\pi)[/itex] a n goes to infinity.
 
  • #5


In a real analysis class you would do something like the following:

Let [tex] f(x) = sin ( \frac {1}{x} ) [/tex]

Let [tex] s_{n} [/tex] be the sequence [tex] \frac {2}{n \pi} [/tex]

[tex] \lim_{n \to \infty}s_{n} = 0 [/tex]

but [tex] (f(s_{n}) )[/tex] is the sequence 1,0,-1,0,1,0,-1,0... which obviously doesn't converge.

Therefore, [tex] \lim_{x \to 0} sin( \frac {1}{x}) [/tex] does not exist.
 
  • #6
tiny-tim
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Hi hermy! :wink:

matt's :smile: is the best way for general questions like this …

it shows you what the difficulty is! :wink:
try drawing it and find out.
 
  • #7
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It would be a bit dull if left and right limits always exist.
Could you give an example of a function where the left/right hand limit doesn't exist at a point (where the function is defined on the set of all real no.s)?
 
  • #8
D H
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f(x) = 0 if x is rational, 1 if it is irrational.
 
  • #9
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Another example is f(x) = 1/x, x [itex]\neq [/itex] 0; f(0) = 0.
This function is defined for all real numbers. The left hand limit, as x approaches zero is negative infinity, while the right hand limit, as x approaches zero, is positive infinity.
 
  • #10
chiro
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Homework Statement



does lim x →0 sin(1/x) exist?

The Attempt at a Solution



I am very new to calculus. From what i have understood till now, limit of a function at a point exists only if the left and right hand limits are equal. What would the graph of sin(1/x) look like? Would the left and right hand limits exist? Do they exist in all cases, for all functions?
In calculus you need to use the idea of limits. Essentially the limit of sin x/x does equal 1 but you have to show it from both sides.

If we consider a left hand limit if sin x/x then we can verify the limit with a table of calculations (just grab a calculator and calculate f(x) sin x / x for values approaching x)

We can also consider the right hand limit also. For the right hand limit we can do the same thing by letting f(x) approach sin x/x.

Now the limit is only valid if and only if the right hand limit equals the left hand limit. So
essentially if this is the case and you find a value for it, then you can prove what that limit
is.
 
  • #11
D H
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The original poster is looking at f(x)=sin(1/x), not g(x)=sin(x)/x.
 
  • #12
matt grime
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Could you give an example of a function where the left/right hand limit doesn't exist at a point (where the function is defined on the set of all real no.s)?
I don't want to give the game away but.... sin(1/x) as x tends to 0 perhaps? I can trivially extend this to a function on the whole of the real line if I want to, if you're worried about it being undefined at zero:

f(x)=sin(1/x) x=/=0
f(0)=0 (or anything else you want)
 

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