Let $f(x,y)=2x-\frac{7}{y}$
Show that $f$ is continuous on the set $E=\{\,(x,y) \, : \, y>0 \, \}$
I did the following:
$\forall \epsilon>0$, $\exists \delta>0$ s.t. $\forall (a,b)\in \mathbb{R^2}$ with $\sqrt{(x-a)^2+(y-b)^2}<\delta$ implies$ |f(x,y)-f(a,b)|<\epsilon$
\begin{align} |f(x,y)-f(a,b)|&= \left|2x-2a+\frac{7}{b}-\frac{7}{y}\right| \\ &\leq 2\big|x-a\big|+\left|\frac{7}{b}-\frac{7}{y}\right| \\ &<2 \delta +\big(\frac{7\delta}{b|y|}\big) \quad \text{(since $b>0$)} \end{align}
What should be the following steps?
\begin{align} &|f(x,y)-f(a,b)|\\ &= \left|2x-2a+\frac{7}{b}-\frac{7}{y}\right| \\ &\leq 2|x-a|+\left|\frac{7}{b}-\frac{7}{y}\right| \\ &<2 \delta +\frac{7\delta}{|b|\,|y|} \end{align}
In the second term $\dfrac{7\delta}{|b|\,|y|}$, we need to keep the infinitesimal $\delta$, and $\dfrac{1}{|b|}$ is just a constant, so we can use $|y-b| < \delta$ to give a lower bound to $|y| > |b| - \delta$. When $\delta$ is sufficiently small ($\delta < \dfrac{|b|}{2}$), we have $|y| > \dfrac{|b|}{2}$. Hence we have
$$ |f(x,y) - f(a,b)| < 2 \delta + 14 \delta / |b| ^2 = 2(1 + 7/|b|^2) \delta.$$
If you want $ |f(x,y) - f(a,b)| < \epsilon$, you just need $2(1+7/|b|^2) \delta \le \epsilon$, or $\delta \le \dfrac{\epsilon}{2(1+7/|b|^2)}$.
To guarantee that $\delta < \dfrac{|b|}{2}$, you just need to set $\delta = \min\left\{\dfrac{|b|}{2}, \dfrac{\epsilon}{2(1+7/|b|^2)}\right\}$.