 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-ii/6/b/solution

L = f^{'} (c)

. Since

x_{n} \leq c \leq y_{n}

and

x_{n} \neq = 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{ϵ ( y _{n} ) ( y _{n} - c ) + ϵ ( x _{n} ) ( c - x _{n} )}{y _{n} - x _{n}} .

(y_{n} - c) / (y_{n} - x_{n})

and

(c - x_{n}) / (y_{n} - x_{n})

are nonnegative and sum to one. Therefore

\frac{f ( y _{n} ) - f ( x _{n} )}{y _{n} - x _{n}} - L \leq max {∣ ϵ (x_{n}) ∣, ∣ ϵ (y_{n}) ∣} ⟶ 0.

Here

ϵ (c) = 0

handles any endpoint of a secant that happens to equal

c

n \to \infty lim \frac{f ( y _{n} ) - f ( x _{n} )}{y _{n} - x _{n}} = f^{'} (c) .

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