3
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ 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 '18 at 0:53
2
$\begingroup$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.