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Use the epsilon-delta definition to prove that $$ \dfrac{\sqrt{x}}{x-1} $$

is continuous on the interval $(0,1)$. I've tried a lot but I just can't seem to see the first step. I don't want the answer just a hint to get me rolling. I know it should be continuous because I looked at the graph, but after $$ |\dfrac{\sqrt{x}}{x-1} - \dfrac{\sqrt{x_0}}{x_0-1}| < \varepsilon $$

I get stuck. I've attempted bringing everything to a common denominator. Subtracting the $x_0$ term to the other side, but to no avail. Any help appreciated

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    $\begingroup$ Have you tried multiplying by $\sqrt {x}+\sqrt {x_0}$ (top and bottom, effectively by $1$) to introduce the term $ |x-x_0|$? $\endgroup$
    – AnyAD
    Sep 24, 2018 at 0:53

1 Answer 1

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Hint: \begin{align*} \left|\frac{\sqrt{x}}{x-1}-\frac{\sqrt{x_0}}{x_0-1}\right|&= \left|\frac{\sqrt{x}}{x-1}+\frac{\sqrt{x}}{x_0-1}-\frac{\sqrt{x}}{x_0-1}-\frac{\sqrt{x_0}}{x_0-1}\right|\\ &\leq \left|\sqrt{x}\right|\left|\frac{1}{x-1}-\frac{1}{x_0-1}\right|+\left|\frac{1}{x_0-1}\right|\left|\sqrt{x}-\sqrt{x_0}\right|. \end{align*}

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