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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 $\displaystyle \delta = \min\left\{\frac{1}{14}, \frac{\varepsilon}{98}\right\}$.

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    Looks good, but in the third paragraph instead of "Then:" you should write "Then $|x-\frac 17| < \delta$ implies". Same after "Also,". – 2017-02-20
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    Just updated it, thanks! – 2017-02-20
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    Some critique: The $\epsilon-\delta$ definition of a limit says that "For *any* $\epsilon >0$, there exists $\delta > 0$..." etc.. You need to state that this works for *any* $\epsilon > 0$. Some texts will say that you need to mention that "Then for all $x \in D(f)$ (the domain of $f(x) = {1\over x}$) satisfying $\left|x-{1\over7}\right| < \delta$...", which might be a better way of writing the above suggestion. And it is much more suitable to state what you were trying to prove, to remind the reader (particularly when proofs are long). So "$\displaystyle \lim_{x\to7}{1\over x} = {1\over7}$". – 2017-02-20
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    It think it looks fine. The details pointed out by @pyrazolam might be implied in your reasoning. For example for *any* $\epsilon>0$ well you produce a $\delta$ regardless so you have a $\delta$ for every $\epsilon>0$. Of course you need to conclude that that $\delta>0$ (because the definition says there should exist a $\delta>0$ - for that you use that $\epsilon>0$). Also that for every $x$ such that $|x-1/7|<\delta$, it's implied without saying that your calculations are valid for any such $x$. – 2017-02-20

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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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    Yes true I didn't think about this! Thank you! – 2017-02-20
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    @danny You're welcome. – 2017-02-20
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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 $x=\frac{1-\beta}7$