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$f\left(x\right)$ is defined on $[a,b]$, differentiable on $\left(a,b\right)$ and has one-sided derivatives at points a and b. How to prove that $f\left(x\right)$ is continuous on $[a,b]$?

The function is continuous on $\left(a,b\right)$ due to necessary condition of differentiability. But I've no idea how to use one-sided derivatives to show continuity at a and b.

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The one-sided derivative at $a$ is $$\lim_{x\rightarrow a+}\frac{f(x)-f(a)}{x-a}$$ which must exists by hypothesis. If $\lim_{x\rightarrow a+}f(x)-f(a)$ did not exist and equal zero, that would provide a contradiction.

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Suppose that $f$ is not continuous at $a$. Then there exists an $\varepsilon > 0$ such for every $n$ there is an $x_n > a$ with $\lvert a - x_n \rvert < 1/n$ and $\lvert f(a) - f(x_n) \rvert \geq \varepsilon$. Then $$\left\lvert \frac{f(a)-f(x_n)}{a-x_n} \right\rvert \geq \frac{\varepsilon}{a-x_n}$$ is unbounded. In particular, this means that $$\frac{f(a)-f(x_n)}{a-x_n}$$ is not convergent.

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