166 Solutions and Hints for Selected Exercises For any x ̸ = 0, we have |xsin(1/x)| = |x||sin(1/x)|≤|x|, which implies −|x|≤ xsin(1/x) ≤|x|. Since limx →0(−| x|)= limx →0 | x| = 0, applying the squeeze theorem yields lim xsin(1/x)= 0. x→0 It now follows that f (x) − f (a) f ′ (0)= lim = lim[xsin(1/x)+ c]= c. x→a x − a x→0 Using Theorem 4.1.2 and the fact that cosx is the derivative of sinx, the derivative of f can be written explicitly as 1 2xsin − cos(1/x)+ c, if x ̸ = 0; f ′ (x)= x c, if x = 0. From the solution, it is important to see that the conclusion remains valid if we replace the function f by 1 xn sin , if x ̸ = 0; g(x)= x 0, if x = 0, where n ≥ 2, n ∈ N. Note that the function h(x)= cx does not play any role in the differentiability of f . We can generalize this problem as follows. Let ϕ be a bounded function on R, i.e., there is M > 0 such that |ϕ(x)|≤ M for all x ∈ R. Defne the function ( xnϕ(1/x), if x ̸ = 0; f (x)= 0, if x = 0, where n ≥ 2, n ∈ N. Then f is differentiable at a = 0. Similar problems: 1. Show that the functions below are differentiable on R: ( 3/2 x cos(1/x), if x ≥ 0; f (x)= 0, if x < 0
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