Using the Definition to Compute the Derivative - Exercise 4


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

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

Answer. We know that

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

Equivalently, we also have

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

Putting these two equations together we will get

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

But

\begin{displaymath}\frac{f(a+h) - f(a)}{h} + \frac{f(a-h) - f(a)}{-h} = \frac{f(a+h) - f(a-h)}{h}\end{displaymath}

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:

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


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