Answer to Question #221516 in Real Analysis for wisdom

Question

17. Let f: I → R, where I is an open interval containing the point c, and let k ∈ R. Prove the following.

(a) f is differentiable at c with f ′(c) = k iff limh→0 [ f (c + h) – f (c)]/h = k.

*(b) If f is differentiable at c with f ′(c) = k, then limh→ 0 [ f (c + h) – f (c – h)]/2h = k.

(c) If f is differentiable at c with f ′(c) = k, then lim n →∞ n[f (c + 1/n) – f (c)] = k.

(d) Find counterexamples to show that the converses of parts (b) and (c) are not true. 

The book is Steven R. Lay, Analysis with an introduction to proof.

Answer 1

Answer:-

We have given $f:I \rightarrow R \ \ \ \ ;$ I is an open interval is differentiable at $c \in (a,b)$

and $f'(c)=k \ \ \ \ ;k\in IR$

then f(x) can be approximated by linear function of x , $\ \ \ \because$ k is a constant

$\therefore$ $f(x)=kx +d \ \ \ \ \ \ \ \ d\in IR$

Now $f(c+h)=k(c+h)+d=kc+kh+d$

$f(c)=kc+d$

$\therefore lim_{h\rightarrow 0}\frac{f(c+h)-f(c)}{h}\\$

$= lim_{h\rightarrow 0} \frac{kc+kh+d-kc-d}{h}$

$=lim_{h\rightarrow 0}\frac{kh}{h}\\ =\boxed{k}$

conversely

$lim_{h\rightarrow 0}\frac{f(c+h)-f(c)}{h}=k$

Consider two points on graph of $f(x) , [c+h,f(c+h)] , [c,f(c)]$

now slope of any line passing through the points is $\frac{f(c+h)-f(c)}{(c+h)-(c)}$

As $lim_{h\rightarrow 0 }$ $\frac{f(c+h)-f(c)}{(c+h)-(c)}$ becomes slope of tangent at point x=c , when is also derivative of function at x=c .

$f'(c)=lim_{\rightarrow 0}\frac{f(c+h)-f(c)}{h}$

$\implies \boxed{f'(c)=k}$

(b)

given , f'(c)=k

$\implies lim_{h\rightarrow 0 }\frac{f(c+h)-f(c)}{h}=k.............(1)\\$

Replace c by c-h

$\implies lim_{h\rightarrow 0 }\frac{f(c-h+h)-f(c-h)}{h}=k.............(1)\\ \implies lim_{h\rightarrow 0 }\frac{f(c)-f(c-h)}{h}=k.............(2)\\$

add (1) and (2)

$\implies\boxed{ lim_{h\rightarrow 0 }\frac{f(c+h)-f(c-h)}{2h}=k}$

(c)

We have given ,

$f'(c)=k$

$\implies lim_{h\rightarrow 0 }\frac{f(c+h)-f(c)}{h}=k$

Replace h by $1\over n$

$\implies As \ h\rightarrow 0 \ \ , n\rightarrow \infin$

$\implies lim_{n\rightarrow 0 }[f(c+\frac{1}{n}-f(c)]=k$

Counter example for converse of (b)

(d)

$f(x) = \begin{cases} 2x+3 &\text{; }x\ge 2 \\ x+5 &\text{l; } x <2 \end{cases}$

$lim _{h\rightarrow 0} \frac{(2+h)-f(2-h)}{2h}$

$=lim _{h\rightarrow 0} \frac{(2(2+h)+3)-f(2-h)}{2h}$

$=lim _{h\rightarrow 0} \frac{4+2h+3-2+h-5}{2h}$

$= \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{3}{2}=\boxed{k}$

now ,

$lim _{h\rightarrow 0} \frac{(c+h)-f(c)}{h}$

$=lim _{h\rightarrow 0} \frac{(2+h)-f(2)}{h}$

$=lim _{h\rightarrow 0} \frac{2(2+h)+3-f(2(2)+3)}{h}$

$=lim _{h\rightarrow 0}\frac{2h}{h}$

$\boxed{2}\ne k \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [k=\frac{3}{2}]$