Disprove the limit $\lim_{x\to 0}\frac{1}{x}=5$ with epsilon-delta

Hint:

What is the negation of $$\forall \varepsilon>0\;\exists \delta>0\;\forall x\;\biggl(\lvert x\rvert<\delta\implies\biggl\lvert\frac1x-5\biggr\rvert<\varepsilon\biggr)?$$

Second hint:

Roughly said, the negation of an implication is a counter-example.

The $\varepsilon-\delta$ definition says:

$\displaystyle\lim_{x\to c} f(x)=L$ means: For all $\varepsilon>0$ there is some $\delta>0$ such that for all $x$ with $0<|x-c|<\delta$ we have $|f(x)-L|<\varepsilon$.

So lets negate this: There is some $\varepsilon>0$ such that for all $\delta>0$ there is some $x$ with $0<|x-c|<\delta$ and $|f(x)-L|\geq \varepsilon$.

Now apply this to $\lim_{x\to 0}\frac{1}{x} \neq 5$.

Let $\varepsilon=420$. Consider any $\delta>0$. Then let $x=\min\{\delta/2,\frac{1}{425}\}$. Then in particular, $\frac{1}{x}\geq 425$. Also note that $0<|x-0|<\delta$ and $|\frac{1}{x}-5|\geq420$ $\square$.

Basically, no matter how close we restrict ourselves to $0$, we can always escape the $\varepsilon$ of room we give ourselves (I picked $420$ just to pick something, you could pick any number really). So no matter how close we are to $0$ the function still blows up there.

Let's assume $\lim_{x\rightarrow 0} 1/x = 5$.

Let $\epsilon = 1$. Then there exist a $\delta$ so the $|x-0| = |x| < \delta \implies |1/x - 5| < \epsilon = 1$. Let $x = \min(\delta/2, 1/6)<\delta$. So $|x| < \delta$ so $|1/x - 5| < 1$. But $x \le 1/6$. So $1/x \ge 6$. So $|1/x - 5| \ge |6-5| = 1$. This is a contradiction.

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More general. For any $\epsilon > 0$ and $\delta > 0$. Let $0 < x < \min(\delta, \frac 1{5+\epsilon})$. Then $|x - 0| < \delta$ and $|1/x - 5| > |1/\frac 1{5+\epsilon} - 5| = |5 + \epsilon - 5| = \epsilon$.

So it is not the case for any $\epsilon > 0$ that there is a $\delta$ so that $|x - 0| < \delta \implies |1/x - 5| < \epsilon$.

So $\lim_{x\rightarrow 0} 1/x \ne 5$.