<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] The only logical error I could find is related to the definition of $\delta$. If you want to use strict inequalities, you should have $\delta<\min\left\{\frac{1}{14},\frac{\varepsilon}{98}\right\}$. Otherwise, at least one of the strict inequalities should be changed (depending on the value of $\varepsilon$). 
Regardless, a good proof if you're still gaining familiarity with $\varepsilon-\delta$ proofs.
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