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Suppose $g:\mathbb{R}\to\mathbb{R}$ is a continuous function which is differentiable at some point $x$. Let $u_n, v_n$ be real sequences with (for all $n$) $u_n\le x \le v_n$, $u_n\neq v_n$ and $u_n, v_n \to x$ as $n\to\infty$. Prove that $$ \lim_{n \to \infty} \frac{g(v_n)-g(u_n)}{v_n-u_n} $$ exists.

I have tried all manners of rearranging the limit to somehow get it in terms of the derivative of $g$ at $x$ but this doesn't really help, I always seem to end up with it in terms of limits which don't necessarily exist. Seeing as there are very few conditions on $g$ I think there should be a relatively simple way of rearranging it but can't spot it for some reason.

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It might be easier to work from the definition.

If $g$ is differentiable at $x$, then for all $\epsilon>0$, there exists some $\delta>0$ such that if $|y-x| < \delta$, then $|g(y)-g(x)-g'(x)(y-x)| < \epsilon |y-x|$.

Now try to bound $|\frac{g(v_n)-g(u_n)}{v_n-u_n} - g'(x)|$, writing $g(v_n)-g(u_n) = g(v_n)-g(x)+g(x)-g(u_n)$, and $g'(x)(v_n-u_n) = g'(x)(v_n-x) + g'(x)(x-u_n) $. Then use the fact that $u_n\le x \le v_n$ to obtain a suitable bound.

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Hint: Try adding and subtracting $g(x)$ in the numerator to get closer to the definition of the derivative.

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Let $\epsilon>0$ be given. As $v_n\to x$, $|g(v_n)-g(x)-(v_n-x)g'(x)|\le \epsilon|v_n-x|=\epsilon(v_n-x)$ for almost all $n$. Similary, $|g(u_n)-g(x)-(u_n-x)g'(x)|\le \epsilon|u_n-x|=\epsilon(x-u_n)$ for almost all $n$. Hence $$\begin{align}&|g(v_n)-g(u_n)-(v_n-u_n)g'(x)|\\=&|(g(v_n)-g(x)-(v_n-x)g'(x))-(g(u_n)-g(x)-(u_n-x)g'(x))|\\\le&|g(v_n)-g(x)-(v_n-x)g'(x)|+|g(u_n)-g(x)-(u_n-x)g'(x)|\\\le&\epsilon(v_n-x)+\epsilon(x-u_n)\\=&\epsilon(v_n-u_n)\\=&\epsilon\,\left|v_n-u_n\right|\end{align}$$ and consequently $$\left|\frac{g(v_n)-g(u_n)}{v_n-u_n}-g'(x)\right|\le \epsilon$$ for almost all $n$.

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Hint: Since $g:\Bbb R\to\Bbb R$ is differentiable at $x$, then $$\begin{align}2g'(x) &= 2\lim_{h\to0}\frac{g(x+h)-g(x)}h\\ &= \lim_{h\to0^+}\left[\frac{g(x+h)-g(x)}h\right]+ \lim_{h\to0^+}\left[\frac{g(x-h)-g(x)}{-h}\right]\\ &= \lim_{h\to0^+}\left[\frac{g(x+h)-g(x)}h\right]+ \lim_{h\to0^+}\left[\frac{g(x)-g(x-h)}h\right]\\ &= \lim_{h\to0^+}\frac{g(x+h)-g(x-h)}h,\end{align}$$ and so $$g'(x) = \lim_{h\to0^+}\frac{g(x+h)-g(x-h)}{2h} = \lim_{h\to0^+}\frac{g(x+h)-g(x-h)}{h-(-h)}.$$ Do you see how this helps you, since $u_n$ and $v_n$ converge to $x$ from below and above (respectively) and $u_n\ne v_n$?

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