165 It is natural to ask whether the function f (x) =x3 is uniformly continuous on R. Following the solution for part (a), we can use xn = 3 rn+ 1 n andyn = 3 √nfor n∈Nto prove that f is not uniformly continuous onR. By a similar method, we can show that the function f (x) =xn, n∈N, n≥2, is not uniformly continuous onR. A more challenging question is to determine whether a polynomial of degree greater than or equal to two is uniformly continuous on R. Exercise 3.5.7. Hint: For part (a) use Theorem 3.5.5. For part (b) prove that the function can be extended to a continuous function on [a,b] and then use Theorem 3.5.5. Exercise 3.5.8. (a) Applying the definition of limit, we findb>asuch that c−1<f (x) <c+1 whenever x >b. Since f is continuous on [a,b], it is bounded on this interval. Therefore, f is bounded on [a,∞). (b) Fix any ε >0, by the definition of limit, we findb>asuch that | f (x)−c| < ε 2 whenever x >b. Since f is continuous on [a,b+1], it is uniformly continuous on this interval. Thus, there exists 0<δ <1 such that | f (u)−f (v)| < ε 2 whenever |u−v| <δ,u,v ∈[a,c+1]. Then we can show that | f (u)−f (v)| <ε whenever |u−v| <δ, u,v ∈[a,∞). (c) Since limx→∞ f (x) =c >f (a), there exists b>asuch that f (x) >f (a) whenever x >b. Thus, inf{f (x) : x ∈[a,∞)}=inf{f (x) : x ∈[a,b]}. The conclusion follows from the Extreme Value Theorem for the function f on [a,b]. SECTION 4.1 Exercise 4.1.10. Use the identity lim n→∞ f (a+1 n) f (a) ! n =lim n→∞ exp(n[ln(f (a+ 1 n ))−ln(f (a)]). Exercise 4.1.11. (a) Using the differentiability of sinx and Theorem 4.1.2, we conclude the function is differentiable at any a̸=0. So, we only need to show the differentiability of the function at a=0. By the definition of the derivative, consider the limit lim x→a f (x)−f (a) x−a =lim x→0 x2sin(1/x)+cx x =lim x→0 [xsin(1/x)+c].
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