= Solution
{c}
Use the <local linear approximation> from part (a), with $L=f'(c)$. Since $x_n\leq c\leq y_n$ and $x_n\ne y_n$, we have $y_n-x_n>0$, and subtraction gives
$$
\frac{f(y_n)-f(x_n)}{y_n-x_n}-L=\frac{\epsilon(y_n)(y_n-c)+\epsilon(x_n)(c-x_n)}{y_n-x_n}.
$$
The coefficients $(y_n-c)/(y_n-x_n)$ and $(c-x_n)/(y_n-x_n)$ are nonnegative and sum to one. Therefore
$$
\left|\frac{f(y_n)-f(x_n)}{y_n-x_n}-L\right|\leq\max\{|\epsilon(x_n)|,|\epsilon(y_n)|\}\longrightarrow0.
$$
Here $\epsilon(c)=0$ handles any endpoint of a secant that happens to equal $c$. Thus <secants straddling a differentiability point> satisfy
$$
\boxed{\lim_{n\to\infty}\frac{f(y_n)-f(x_n)}{y_n-x_n}=f'(c).}
$$
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