<s>[INST] Epsilon-Delta: Prove $\frac{1}{x} \rightarrow 7$ as $x \rightarrow \frac{1}{7}$ 
Prove that $\displaystyle\frac{1}{x} \rightarrow 7$ as $\displaystyle x \rightarrow \frac{1}{7}$.

I need to show this with an $\epsilon-\delta$ argument. Still figuring these types of proofs out though, so I could use some tips/critiques of my proof, if it is correct at all. 
It might not be so clear, but I use the fact that $\displaystyle\left|x - \frac{1}{7}\right| < \delta$ several times in the proof. 
For $\varepsilon > 0$, let $\displaystyle\delta = \min\left\{\frac{1}{14}, \frac{\varepsilon}{98}\right\}$. Then $\displaystyle \left|x - \frac{1}{7}\right| < \delta$ implies:
$$\left|\frac{1}{7}\right| = \left|\left(-x + \frac{1}{7}\right) + x\right| \leq \left|x - \frac{1}{7}\right| + \left|x\right| < \frac{1}{14} + |x|,$$ and so $\displaystyle |x| > \frac{1}{14}$.
Also, $\displaystyle \left|x - \frac{1}{7}\right| < \delta$ implies:
$$\left|\frac{1}{x} - 7\right| = \left|\frac{1-7x}{x}\right| = 7\frac{\left|x - \frac{1}{7}\right|}{|x|} < 98\left|x - \frac{1}{7}\right| < \frac{98\varepsilon}{98} = \varepsilon.$$
Thus for $\varepsilon > 0$, $\displaystyle\left|\frac{1}{x} - 7\right| < \varepsilon$ if $\displaystyle\left|x - \frac{1}{7}\right| < \delta$, for  $\displaystyle \delta = \min\left\{\frac{1}{14}, \frac{\varepsilon}{98}\right\}$.
 [/INST] for $\beta \gt 0$ write:
$$
x = \frac{1+\beta}7
$$
then
$$
\begin{align}
7-\frac1x\ & \le 7\bigg(1-\frac1{1+\beta}\bigg) \\[14pt]
& = \frac{7\beta}{\beta+1}\\[14pt]
& \le 7\beta
\end{align}
$$
thus:
$$
x-\frac1{7} \le \frac{\beta}7 \Rightarrow 7 -\frac1x\le 7\beta
$$
with this established, the $\epsilon-\delta$ argument is straightforward. a minor adjustment provides the argument for the case $x=\frac{1-\beta}7$ 
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