166 Solutions and Hints for Selected Exercises For any x̸ =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 x→0 xsin(1/x) =0. It now follows that f ′(0) =lim x→a f (x)−f (a) x−a =lim x→0 [xsin(1/x)+c] =c. Using Theorem 4.1.2 and the fact that cosx is the derivative of sinx, the derivative of f can be written explicitly as f ′(x) = 2xsin 1 x − cos(1/x)+c, if x̸ =0; c, if x =0. From the solution, it is important to see that the conclusion remains valid if we replace the function f by g(x) = xnsin 1 x , if x̸ =0; 0, if x =0, where n≥2, n∈N. Note that the functionh(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)| ≤Mfor all x ∈R. Define the function f (x) =( xnϕ(1/x), if x̸ =0; 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: f (x) =( x3/2cos(1/x), if x ≥0; 0, if x <0
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