Question

a) Let D be a nonempty subset of R, bounded above. Is M = sup(D) a limit point of D? Explain.

b) Let the function g : R → R be bounded. Define f(x) = 1 + 4x + x^2(g(x)). Prove that

f(0) = 1 and f'(0) = 4. (You may not assume g is differentiable.)

c) Suppose that the function f : R → R is differentiable and that there is a bounded sequence {xn} with xn doesnt= xm if n doesnt= m, such that f(xn) = 0 for every index n. Show that there is a point x0 at which f(x0) = 0 and f'(x0) = 0. (Hint: Use Bolzano-Weierstrass Theorem.)

Answer 1

a)

let us an example;

D = {1,2,3,4}

D is non empty

D is bounded above

Supremum of D = 4

'4' is not a limit point of D

for $\in$ = 1/2

4 $\in$ (4 - 1/2, 4+ 1/2)$\notin$ 4

therefore is not limit point

supremum of D is not a limit point for D

b)

f(x) = 1 +4x + x ² (gx)

if x = 0

f(x) = 1 + 0 +0

f(x) = 1

therefore g is bounded

so g * 0 = 0

f(0) = 1

lim ( f(x) - f(0))/(x - 0)

x $\to$ 0

= (1 + 4x + x² - 1)/ x

= lim 4 + x

x$\to$0

= 4

thus lim (f(x) - f(0))/(x - 0) = 4 which is finite value

x$\to$ 0

so we have f'(0) = 2

c)

Let f: IR$\to$ IR be a function, then lim f(x) = $\rho$ $\isin$ IR if and only if for any sequence (fn) in IR with n$\to$a, f(m/ $\to$ $\rho$ n$\to$ a

Given f : IR$\to$IR is differentiable and x(n) $\subset$ IR with x(n) $\notin$ x(m) if n $\notin$ m such that f(xn) = 0 $\forall$ n $\isin$ IN and xn is bounded sequence. Hence by Bolzano-Weierstrass theorem, $\exist$ a subsequence (xn_R) of (xn) such that (xn_R) convergent and converges to x_o $\in$ IR

Now f'(x₀) = lim (f(x) - f(x₀))/(x - x₀₎ exist

x$\to$ x₀

we know xn_R$\to$ x₀ and hence;

lim (f(xn_R) - f(x₀))/ (xn_R - x₀)

R$\to$ $\empty$

lim (0 - 0)/ (xn_R - x₀)

R$\to$ 0

= 0

Hence f'(x₀) = lim (f(x) - f(x₀))/ (x - x₀)

x$\to$x₀

= lim (f(xn_R) - f(x₀)) / (xn_R - x₀)

R$\to\empty$

= 0

Therefore f'(x₀) = 0