Using the Definition to Compute the Derivative - Exercise 4

Exercise 4. Show that if f'(a) exists, then we have

$$\lim_{h \rightarrow 0} \frac{f(a+h) - f(a-h)}{h} = 2 f'(a)\;.$$

Answer. We know that

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

Equivalently, we also have

$$\lim_{h \rightarrow 0} \frac{f(a-h) - f(a)}{-h} = f'(a)\;.$$

Putting these two equations together we will get

$$\lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h} + \frac{f(a-h) - f(a)}{-h} = 2f'(a)\;.$$

But

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

which gives the desired relation.

Note that this formula is for example used by calculators to approximate f'(a) especially when f(x) is known for values of x near a.

In the same spirit as the example above, one can prove the following formula which involves the second derivative:

$$\lim_{h \rightarrow 0} \frac{f(a+h) - 2f(a) + f(a-h)}{h^2} = f''(a)\;.$$

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