166 Solutions and Hints for Selected Exercises For anyx̸ =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 onR: f (x)=( x3/2cos(1/x), if x ≥0; 0, if x <0
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