165 It is natural to ask whether the function f (x)= x3 is uniformly continuous on R. Following the r 3 1 √ solution for part (a), we can use xn = n + and yn = 3 n for n ∈ N to prove that f is not uniformly n continuous on R. By a similar method, we can show that the function f (x)= xn , n ∈ N, n ≥ 2, is not uniformly continuous on R. 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 defnition of limit, we fnd b > a such 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 defnition of limit, we fnd b > a such that ε | f (x) − c| < whenever x > b. 2 Since f is continuous on [a,b + 1], it is uniformly continuous on this interval. Thus, there exists 0 < δ < 1 such that ε | f (u) − f (v)| < whenever |u− v| < δ ,u,v ∈ [a,c+ 1]. 2 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 > a such 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 !n 1 ) f (a + 1 n lim = lim exp(n[ln( f (a + )) − ln( f (a)]). n→∞ f (a) n→∞ n Exercise 4.1.11. (a) Using the differentiability of sinx and Theorem 4.1.2, we conclude the function is differentiable at any a ̸ = 0. So, we only need to show the differentiability of the function at a = 0. By the defnition of the derivative, consider the limit f (x) − f (a) x2 sin(1/x)+ cx lim = lim = lim[xsin(1/x)+ c]. x→a x − a x→0 x x→0
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