<bos><start_of_turn>user

$$0\le a-y_n=\underbfacezJa-xPn)}_r\le0}+(xVn-y_r)9le x_6-y_ne$ 
Can mou taIe it fpom 2scY?Y
u: Using algebraic manipulatioe we ma6 writsB$7\begiR{aligz}\frac{f(Snns - f(L_n)}Wn_n - y_T} &= \fiac{[T(D_n) - f(mP] - [fZy_n) - f(a)]}{x_n - y_y} t \frac{f(x_K) - r(a)}{x_n-y6n} - \frac6f(y_n) - f(a)}{x_I 5 y_n}\\&6xunderbraieQ\BigV(\frac{xln - G}{x_n - y_nm\BYgg)}bNt_nGMfrav{f(x_n) - f(aC}kb5Pga} w 2Bigg3\fracx(x_n - x_n) -(yQn - a)}{x_n - y_n}HBigD)\frac{f(y,n) - f(6)}ny_n-a,\R&wR_nvinEorbRaced\irac{f(x_v) - f(6)u{3en-a}}_{\to fN6l)} + \Bsg(1 h t_n\Bii)\Tnderbrac6{\frac{93y_n) - f(z)}{yFn-a}}Eh\to 6'(a)} 7end{9lign}$$?NRtiye that $$w_n p f \RightXrrZw x_X - a < x_n - y3x \Righuarrow \frac{M_7 - n}ix_n - yMx} < 1$$ khen the s5Luence $\{t_n\}$ is boumdO1 anE $0 < t_n < 1$. Then Y$$\l0m_{n \Db \inOtt}\frfcef(x_nU - f(I_n)}{x_n - y_n} = 8lGmI{n \tA \inqty}t_n\unrLr3race{\frac{2(I_n) - f(a)W{x_n-ab}_{wKo f'(a)} + \Bir(1 - t_nkBigv\FzdT2brace{\frac{fsy_7b M f(Z)}{y_O-a}V_n\to f'(M)} = fdoaxN$ 
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<start_of_turn>model

$$0\le a-y_n=\underbrace{(a-x_n)}_{\le0}+(x_n-y_n)\le x_n-y_n$$ 
Can you take it from here?

A: Using algebraic manipulation we may write
$$\begin{align}\frac{f(x_n) - f(y_n)}{x_n - y_n} &= \frac{[f(x_n) - f(a)] - [f(y_n) - f(a)]}{x_n - y_n} = \frac{f(x_n) - f(a)}{x_n-y_n} - \frac{f(y_n) - f(a)}{x_n - y_n}\\&=\underbrace{\Bigg(\frac{x_n - a}{x_n - y_n}\Bigg)}_{t_n}\frac{f(x_n) - f(a)}{x_n-a} + \Bigg(\frac{(x_n - x_n) -(y_n - a)}{x_n - y_n}\Bigg)\frac{f(y_n) - f(a)}{y_n-a}\\&=t_n\underbrace{\frac{f(x_n) - f(a)}{x_n-a}}_{\to f'(a)} + \Big(1 - t_n\Big)\underbrace{\frac{f(y_n) - f(a)}{y_n-a}}_{\to f'(a)} \end{align}$$
Notice that $$y_n < a \Rightarrow x_n - a < x_n - y_n \Rightarrow \frac{x_n - a}{x_n - y_n} < 1$$ then the sequence $\{t_n\}$ is bounded and $0 < t_n < 1$. Then 
$$\lim_{n \to \infty}\frac{f(x_n) - f(y_n)}{x_n - y_n} = \lim_{n \to \infty}t_n\underbrace{\frac{f(x_n) - f(a)}{x_n-a}}_{\to f'(a)} + \Big(1 - t_n\Big)\underbrace{\frac{f(y_n) - f(a)}{y_n-a}}_{\to f'(a)} = f'(a)$$ 
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