Let $f(x)$ be differentiable at $x = a$...

Question

...and define $F(\Delta x) = \frac{f(a + \Delta x) - f(a - \Delta x)}{2\Delta x}$, $x= a+ \Delta x$. What is $\lim_{\Delta x \rightarrow 0} F(\Delta x)$?

So far I have this:

$\lim_{\Delta x \to 0} F(\Delta x) = \lim_{\Delta x \to 0} \frac{f(a + \Delta x) - f(a - \Delta x)}{2\Delta x} = \lim_{x \to a} \frac{f(x) - f(2a - x)}{2(x-a)}$. I am not sure where to go from here; I think the final answer is $0$.

Also, as an aside, what does $F$ mean geometrically? I am having a hard time understanding what the function entails.

Answer 1

Let $x_0= x- \Delta x$. Then $x= x_0+ \Delta x$ and $x+ \Delta x= x_0+ 2\Delta x$ so that $\frac{f(x+ \Delta x)- f(x- \Delta x)}{2\Delta x}= \frac{f(x_0+ 2\Delta x)-f(x_0)}{2\Delta x}$. Taking $h= 2\Delta x$ that becomes $\frac{f(x_0+ h)- f(x_0)}{h}$ and the limit of that as $\Delta x$ (and so h) goes to 0 is the derivative at $x_0$. But since $\Delta x$ is going to 0, $x_0= x- \Delta x$ goes to x. That is, this is simply another way of finding the derivative of f(x). Geometrically, while the usual formula for the derivative take the "secant" from x to $x+ \Delta x$ which, in the limit, becomes the tangent line, this formula takes the secant from two points at equal distance from x. In the limit, that also becomes the tangent line.

Answer 2

Hint: Write $$f(a+\Delta x)-f(a-\Delta x)=\big[\,f(a)-f(a-\Delta x)\big]-\big[\,f(a)-f(a+\Delta x)\big]$$ to relate $\lim_{\Delta x\to0}F(\Delta x)$ to $f'(a)$.

Geometrically, just draw a picture like you would for the usual limit that defines the derivative: $f(a+\Delta x)-f(a-\Delta x)$ is the (signed) height of the line segment that joins $(a-\Delta x,f(a-\Delta x))$ with $(a+\Delta x,f(a+\Delta x))$, and $2\Delta x$ its width.

Answer 3

Let $h = \Delta x$, and note that $a = x - h$. Then $F(h)$ is

$$ F(h)= \frac{f(x) - f(x - 2h)}{2h} $$

Use $h' = -h/2$ instead of $h$, for mere psychological reasons. The limit asked becomes

$$ lim_{h\to 0} F(h) = lim_{h'\to 0} \frac{f(x+h')-f(x)}{h'} = f'(x) $$

See that $F(h)$ is the slope of the secant $s(x)$ passing through $f(x+h)$ and $f(x-h)$. $$ s(x) = f(x-h) + \frac{f(x+h) - f(x-h)}{(x+h)-(x-h)}x $$

When the limit is taken, this becomes $f'(x)$, as expected, because the secant becomes the tangent when $h\to0$.