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MTH 173
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** Question Form_labelMessages **
question about final exam
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I was preparing for the final exam and I encountered some problems.
I was having problem with questions that talk about continuity and differentiability of sin(1/x) at x = 0.
I know the fact that sin(1/x) assumes all the values from -1 to 1 in the vicinity of x = 0. But does that make the function continuous at x = 0? and if so how do i talk about the differenitability of the function at x = 0. My guess for the same is not differentiable..
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Within any interval containing x = 0 the values of 1/x take values that span an infinity interval (for example within the x interval (-.0001, .0001) the values of 1/x span the intervals (-10 000, infinity) and (10 000, infinity)), always containing at least one interval of 2 pi (in fact, an infinite number of such intervals, but one is enough for our purposes).
So on any interval containing x = 0, sin(1/x) takes all values between -1 and 1.
sin(1/x) is not defined at x = 0, since division by 0 is undefined. However this doesn't mean we can't choose some value and assign it to the function at x = 0.
However, since sin(1/x) takes all values between -1 and 1 on any interval about x = 0, its values cannot be confined to any smaller neighborhood of our chosen value. So sin(1/x) is not defined at x = 0.
In order to be differentiable at x = 0 we would have to show that the limit of the expression
(f(x + `dx) - f(x)) / `dx
exists as `dx -> 0 . We cannot do so, though, because at x = 0 the expression f(x) cannot be defined.
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All this can be rephrased more rigorously in terms of epsilons and deltas.
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