Let E С M.m, x 6 E and a 6 MP. Consider a function f : E \ {x} -» RP (or4 a function f : E -» W). We say that f (у) ^ а as у —»• x through values of у 6 E if, given e > 0, we can find a S(e, x) > 0 such that, whenever у 6 E and 0 < ||x — y|| < 5(e,x), we have
||f(x) -a|| <e. (We give a slightly more general definition in Exercise K.23.)Exercise 4.35. Let E С W71, x 6 E and a 6 W. Consider a function f : E \ {x} -» RP. De/me i : E ^ W by f (y) = f (у) г/ у € £7 \ {x} and f (x) = a. STioiu that f (y) —>• a as у —»• x through values of у £ E if and only if f is continuous at x.
Exercise 4.36. Д/iter looking at your solution of Lemma 4-29, state and prove the corresponding results for limits.
Exercise 4.37. [Ln this exercise you should start from Definition 1.Щ Let U be an open set in Ж. Show that a function f : U —> Ж is differentiable at t 6 U with derivative f'(t) if and only if
f(t + h)-f(t) ^
as h —»• 0 (through values of h with t + h 6 U).
Exercise 4.38. Ln Chapter 6 we approach the properties of differentiation in a more general manner. However the reader will probably already have met results like the following which can be proved using Exercises 4-36 and 4.31.
(i) Lf /, g : (a, b) —»• Ж are differentiable at x 6 (a, b), then so is the sum f + g and we have (/ + g)'(x) = f'(x) + g'(x).
(ii) Lf f, g : (a,b) —»• M are differentiable at x 6 (a,b), then so is the product f x g and we have (/ x g)'(x) = f'(x)g(x) + f(x)g'(x). [Hint: f(x+h)g(x+h)-f(x)g(x) = (f(x+h)-f(x))g(x+h)+f(x)(g(x+h)-g(x)).]
(Hi) Lf f : (a, b) —»• M is nowhere zero and f is differentiable at x 6 (a, b), then so is 1// and we have (l/f)'(x) = —f'(x)/f(x)2.
(iv) State accurately and prove a result along the lines of (ii) and (Hi) dealing with the derivative of f/g.
(v) Lf с 6 Ж, с ф 0 and f : Ж —> Ж is differentiable at x, show that the function fc defined by fc(t) = f(ci) [t 6 Ж] is differentiable at с_1ж and we have f'c(c~1x) = cf'(x). What happens if с = О?
(vi) Use part (ii) and induction on n to show that if rn{x) = xn, then rn is everywhere differentiable with r'n(x) = nrn-\{x) [n > 1]. Hence show that every polynomial is everywhere differentiable. Lf P and Q are polynomials and Q(t) ф 0 for all t 6 (a, b) show that P/Q is everywhere differentiable on (a, b).