Question

In Spivak's Calculus, 4th Edition, page 96, he shows that sin (1/x) has no limit as x approaches 0 with the following:

...It is false that f approaches 0 near 0. This amounts to saying that it is not true for every number ε>0 that we can get |f(x)-0| < ε by choosing x sufficiently small, and x ≠ 0. To find this we simply have to find one ε > 0 for which the condition |f(x)-0| < ε cannot be guaranteed, no matter how small we require |x| to be. In fact, ε = 1/2 will do: it is impossible to ensure that |f(x) < 1/2| no matter how small we require |x| to be; for if A is any interval containing 0, there is some number x = 1/(1/2ᴨ + 2nᴨ) which is in this interval, and for this x we have f(x) = 1.

My question is:

I don't understand how selecting ε leads to the interval A and what is the n in x = 1/(1/2ᴨ + 2nᴨ)?

Answer 1

I'm not sure he is asserting that epsilon leads to A, it sounds like he is just making an inference in the form of P then Q, with "if A is any interval containing 0" being the antecedent (P).

So, given an interval A which includes 0, there is always some shorter period you can make the sine wave (by dividing it by 2, regardless of how many complete wave periods [+2npi] you arbitrarily shorten the period. To me, in plain english it looks like he is saying that it just oscillates tighter and tighter without limit as x approaches zero.

Answer 2

Thanks, Ajk...any thoughts on how he got x = 1/(1/2ᴨ + 2nᴨ) so that f(x) = 1 ?

Answer 3

Isn't 1 the abs value of the maximum distance from the x axis that sin can reach without amplification? That would be my intuition.

Answer 4

Or, just in plain english (my preference), the range of sin(|x|) is (0,1]

Answer 5

same for sin(1/x), unless I'm mistaken.

Answer 6

sin(|x|) is still between -1 and 1 ...

|sin(x)| is between 0 and 1

Answer 7

Ah yeah, good catch. ty.

Answer 8

yw, im sure it was a typo :)