Let $ a < b $. Let $ f : (a,b) \rightarrow \mathbb{R} $ be differentiable. Let $ f' :(a,b) \rightarrow \mathbb{R} $ be continuous.
Let $x_0 \in (a,b)$. $ (u_n)$ , $(v_n) $ are two zero sequences. Moreover $u_n > v_n $ and $u_n + x_0 $ and $ v_n + x_0 \in (a,b)$. Show $$ \lim_{n \rightarrow \infty} \frac{ f(x_0 + u_n) - f(x_0 +v_n)}{u_n -v_n} = f'(x_0) .$$
My work: if I consider the limit I would have $\frac{0}{0}$. So I think that I have to use L'Hospital. Is this true? But how do I know that this is $f'(x_0)$? By the way this our difference quotient : $ \lim_{h \rightarrow 0} \frac{ f(x_0 + h) - f(x_0)}{h} $
Hint. By the Mean value theorem, there is $t_n$ between $x_0 + u_n$ and $x_0 + v_n$ such that $$\frac{ f(x_0 + u_n) - f(x_0 +v_n)}{u_n -v_n}=f'(t_n).$$ Now use the fact that $x_0+u_n\to x_0$, $x_0+v_n\to x_0$ and $f'$ is continuous at $x_0$