3.1.6. Suppose that $f:(a, b) \rightarrow \mathbf{R}$ is differentiable, $a<c<b$, and $f^{\prime}(c+)$ and $f^{\prime}(c-)$ exist. Show that $f^{\prime}(c+)=f^{\prime}(c)=f^{\prime}(c-) .$
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To show that $f^{\prime}(c)=f^{\prime}(c+)$, let $x_{n} \rightarrow c+.$ Then, for each $n \geq 1$, there is a $y_{n}$ between $c$ and $x_{n}$, such that $f\left(x_{n}\right)-f(c)=f^{\prime}\left(y_{n}\right)\left(x_{n}-c\right) .$ Since $x_{n} \rightarrow c+$ $y_{n} \rightarrow c+.$ It follows that $f^{\prime}\left(y_{n}\right) \rightarrow f^{\prime}(c+) .$ Hence, $\left(f\left(x_{n}\right)-f(c)\right) /\left(x_{n}-c\right) \rightarrow$
$f^{\prime}(c+)$, i.e., $f^{\prime}(c)=f^{\prime}(c+) .$ Similarly, for $f^{\prime}(c-)$.
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