Exercise 5.9

Question

Exercise 9: Let $f$ be a continuous real function on $\mathbf R$, of which it is known that $f'(x)$ exists for all $x
eq 0$ and that $f'(3)ightarrow 3$ as 
    $xightarrow 0$. Does it follow that $f'(0)$ exists?

ghostofgarborg · Accepted Answer

(	extit{Matt ``Frito'' Lundy}) \
    By definition
    $$f'(0) = \lim_{t 	o 0}\frac{f(t) - f(0)}{t}.$$
    Because $f$ is continuous on $\mathbf R$ and differentiable on $\mathbf R \setminus 0$, 
    ``the'' mean value theorem says for any $t \in \mathbf R \setminus 0$
    , there exists an $x_t \in (0,t)$ (or $(t,0)$ if $t$ is negative)
    such that
    $$\frac{f(t) - f(0)}{t} = f'(x_t).$$
    So we have:
    $$f'(0) = \lim_{t 	o 0}\frac{f(t) - f(0)}{t} = \lim_{t 	o 0} f'(x_t) = 3.$$
    Because $x_t 	o 0$ as $t 	o 0$.
    So $f'$ exists at $0$, and is $3$.

Exercise 5.9

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Exercise 5.9

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