If $f'(t)$ exists then $\lim_{h,k\to0}\frac{f(t+h)-f(t-k)}{h+k}=f'(t)$

Question

Consider $f'$ continuous...

I'm tried to prove it...with Taylor's expansion:

$f(t+h)=f(t)+hf'(t)+h\phi(t,h)$

$f(t-k)=f(t)-kf'(t)-k\phi(t,-k)$

Then:

$\frac{f(t+h)-f(t-k)}{h+k}=f'(t)+\frac{h}{h+k}\phi(t,h)+\frac{k}{h+k}\phi(t,-k)$

I don't know how I can prove $\lim_{h,k\to0}\frac{h}{h+k}\phi(t,h)=0$

Answer 1

If $f'$ is continuous, by the MVT, we have $\frac{f(t + h) - f(t - k)}{(t + h) - (t - k)} = f'(\xi)$ for some $\xi$ between $t+h$ and $t - k$. As $h, k \to 0$, we have $\xi \to t$.

What if $f'$ is not continuous?

A counterexample

Let $f: (a, b) \to \mathbb{R}$ be differentiable. Let $t \in (a, b)$. Can we say $\lim_{h, k\to 0} \frac{f(t + h) - f(t - k)}{h + k} = f'(t)$?

No. $\lim_{h, k\to 0} \frac{f(t + h) - f(t - k)}{h + k}$ may not exist. Let $$f(x) = \left\{\begin{array}{cc} x^2\sin\frac{1}{x} & x\ne 0 \\ 0 & x = 0. \end{array} \right.$$ Let $t = 0$. $f$ is differentiable on $\mathbb{R}$ and its derivative is given by $$f'(x) = \left\{\begin{array}{cc} 2x\sin\frac{1}{x} - \cos \frac{1}{x} & x\ne 0 \\[5pt] 0 & x = 0. \end{array} \right.$$

Let us prove that $\lim_{h, k\to 0} \frac{f(h) - f( - k)}{h + k}$ does not exist. Consider $h_n = \frac{1}{n}$ and $k_n = -\frac{1}{n} + \frac{1}{n^2}$. Then as $n \to \infty$, we have $(h_n, k_n) \to (0, 0)$. However, we have \begin{align} &\frac{f(h_n) - f( - k_n)}{h_n + k_n} \\ =\ & \sin n - \sin\left(n + 1 + \frac{1}{n-1}\right) + \frac{2n - 1}{n^2}\sin \frac{n^2}{n-1}\\ =\ & -2\cos\left(n + \frac{1}{2} + \frac{1}{2n - 2}\right) \sin\left(\frac{1}{2} + \frac{1}{2n - 2}\right) + \frac{2n - 1}{n^2}\sin \frac{n^2}{n-1}. \end{align} Clearly, it does not converge to zero as $n\to \infty$.

Discussion

However, the following is true:

Suppose that $f$ is differentiable at $x$. Then $$f'(x) = \lim_{h\to 0^{+},\ k\to 0^{+}} \frac{f(x+h) - f(x-k)}{h+k}.$$ Calculus by Spivak, 3rd Ed., page 164, question 22(b).

See: Prove that $f'(x) = \lim_{h\to 0^+ \\k\to 0^+} \frac{f(x+h) - f(x-k)}{h+k}$

Answer 2

You went wrong way. It is simpler:

$$\frac{f(t+h)-f(t-k)}{h+k}=\frac{f(t+h)-f(t)+f(t)-f(t-k)}{h+k}=$$

$$h\frac{f(t+h)-f(t)}{h(h+k)}+k\frac{f(t)-f(t-k)}{k(h+k)}$$

And now

$$\lim_{h\to 0}\frac{f(t+h)-f(t)}{h}=\lim_{k\to 0}\frac{f(t)-f(t-k)}{k}=f'(t)$$

making

$$\lim_{h,k\to 0} \left ( h \frac{f(t+h)-f(t)}{h(h+k)}+k\frac{f(t)-f(t-k)}{k(h+k)} \right ) =$$ $$f'(t)\lim_{h,k\to 0}\left ( \frac{h}{h+k}+\frac{k}{h+k} \right ) =f'(t)\lim_{h,k\to 0}\frac{h+k}{h+k}=f'(t)$$

Clarification:

To be totally precise, even for untrained eyes, this should be derived in the opposite direction.

If $\lim a$ exists and $\lim b$ exists then $\lim ab$ exists.

$$\lim_{h,k\to 0}\frac{h+k}{h+k}=1$$

$$\lim_{h\to 0}\frac{f(t+h)-f(t)}{h}=\lim_{k\to 0}\frac{f(t)-f(t-k)}{k}=f'(t)$$

So

$$\lim_{h,k\to 0}f'(t)\frac{h+k}{h+k}=f'(t)$$

$$\lim_{h,k\to 0} \left ( \frac{hf'(t)}{h(h+k)}+\frac{kf'(t)}{k(h+k)} \right ) = f'(t)$$

and now expand $f'(t)$ in two different but equivalent ways and there you go.

The entire trick here is dealing with expressions that have predefined limits.