Answer to Question #221651 in Calculus for Anuj

Question

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

a) By the definition, $f$ is differentiable at $c$ with $f '(c)=k$ iff

$f(x)=f(c)+k(x-c)+o(|x-c|),\,x\to c$. Then, by denoting $x-c$ as h, we have:

$f$ is differentiable at $c$ with $f '(c)=k$ iff

$f(c+h)-f(c)=kh+o(|h|),\,h\to 0$ iff

$\frac{f(c+h)-f(c)}{h}=k+\frac{o(|h|)}{h}=k+o(1)$ iff

$\lim\limits_{h\to 0}\frac{f(c+h)-f(c)}{h}=k$.

b) Using the result from a), we have

$\lim\limits_{h\to 0}\frac{f(c+h)-f(c)}{h}=k$ and $\lim\limits_{h\to 0}\frac{f(c)-f(c-h)}{h}=\lim\limits_{h\to 0}\frac{f(c-h)-f(c)}{-h}=\lim\limits_{h\to 0}\frac{f(c+h)-f(c)}{h}=k$.

Therefore,

$\lim\limits_{h\to 0}\frac{f(c+h)-f(c-h)}{2h}=\frac{1}{2}\lim\limits_{h\to 0}(\frac{f(c+h)-f(c)}{h}+\frac{f(c)-f(c-h)}{h})=\frac{1}{2}(k+k)=k$.

c) If $f$ is differentiable at $c$ with $f'(c)=k$ , then $\lim\limits_{h\to 0}\frac{f(c+h)-f(c)}{h}=k$. That is, for all $\varepsilon>0$ there exists $\delta>0$ such that for all $h\in(-\delta,\delta)$ we have $|\frac{f(c+h)-f(c)}{h}-k|<\varepsilon$.

Put $N=[1/\delta]+1>1/\delta$. For all $n\in\mathbb{N}$, if $n>N$ then $1/n<1/N<\delta$ and hence,

$|\frac{f(c+1/n)-f(c)}{1/n}-k|<\varepsilon$, that is, $|n({f(c+\frac{1}{n})-f(c)})-k|<\varepsilon$.

This means that $\lim\limits_{n\to +\infty}n({f(c+\frac{1}{n})-f(c)})=k$.

d) Let $f(x)=\cos\frac{2\pi}{x}$, if $x\ne 0$, else $f(x)=1$. Consider $c=0$.

Then

$\lim\limits_{h\to 0}\frac{f(c+h)-f(c-h)}{2h}=\lim\limits_{h\to 0}\frac{\cos\frac{2\pi}{h}-\cos\frac{2\pi}{(-h)}}{2h}=0$ and

$\lim\limits_{n\to +\infty}n({f(c+\frac{1}{n})-f(c)})=\lim\limits_{n\to +\infty}n({\cos 2\pi n-1})=0$

We see that both conditions b) and c) are met, but $f(x)$ is not differentiable at $x=0$. It is not even continuous there, since the sequence $f(c+\frac{1}{2n})=\cos\pi n=(-1)^n$ has no limit.