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Consider the function f(x)=sin(1/x)

Find a sequence of x-values that approach 0 such that

(1) sin (1/x)=0 {Hint: Use the fact that sin(pi) = sin(2pi)=sin(3pi)=...=sin(npi)=0}
(2) sin (1/x)=1 {Hint: Use the fact that sin(npi)/2)=1 if n= 1,5,9...}
(3) sin (1/x)=-1
(4) Explain why your answers to any of parts(1-3) show that lim X->0 sin(1/x) does not exist.

Is sin sin (1/x)=0 and sin (1/x)=-1 does not exist.

What is sin (1/x)=1 then.

  • Calculus -

    plz answer

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