Exercise 5.2.6

Question

Let $g$ be defined on an interval $A$, and let $c \in A$.
  \begin{enumerate}[label=(\alph*)] 
  \item Explain why $g^{\prime}(c)$ in Definition 5.2.1 could have been given by
    $$
    g^{\prime}(c)=\lim _{h \rightarrow 0} \frac{g(c+h)-g(c)}{h} .
    $$
  \item Assume $A$ is open. If $g$ is differentiable at $c \in A$, show
    $$
    g^{\prime}(c)=\lim _{h \rightarrow 0} \frac{g(c+h)-g(c-h)}{2 h} .
    $$
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item This is just a change of variable from the normal definition, set $h = (x - c)$ to get
    $$\lim_{x 	o c} \frac{g(x) - g(c)}{x-c} = \lim_{h 	o 0} \frac{g(c+h) - g(c)}{h}$$
  \item Some basic algebra
    $$\lim_{h 	o 0} \frac{g(c+h) - g(c) + g(c) - g(c-h)}{2h} = \frac 12\left(\lim_{h 	o 0} \frac{g(c+h)-g(c)}{h} + \lim_{h 	o 0} \frac{g(c) - g(c-h)}{h}ight)$$
    The first term is clearly $g'(c)$ and the second is $g'(c)$ with the substitution $h = c-x$ since
    $$
    g'(c) = \lim_{x 	o c} \frac{g(c) - g(x)}{c-x}
    $$
    Thus the whole expression is $g'(c)$.
   \end{enumerate}

Exercise 5.2.6

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Exercise 5.2.6

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