If $f'(t)$ exists then $\lim_{h,k\to0}\frac{f(t+h)-f(t-k)}{h+k}=f'(t)$ 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 $\lim_{h,k\to0}\frac{h}{h+k}\phi(t,h)=0$
 A: 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 $f'(x) = \lim_{h\to 0^+ \\k\to 0^+} \frac{f(x+h) - f(x-k)}{h+k}$
A: 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.
