Prove $\lim_{x\to\infty} \frac{1}{x} = 1$ is false $$\lim_{x\to\infty} \frac{1}{x} = 1$$
Given $\epsilon > 0$
$$\left|\frac{1}{x} - 1\right| < \epsilon.$$
Rewrite it as
$$-\epsilon < \frac{1}{x} - 1 < \epsilon$$
$$-\epsilon + 1< \frac{1}{x} < \epsilon + 1$$
If epsilon is very small, then on both sides we are getting value close to $1$, but the function gets closer to zero, hence both sides false.
If $\epsilon$ is big, then on the right side we are getting big positive value, but with $ n \in (0,1)$ the function gets bigger too. Hence right side fails.
Is this  a sound proof? And if yes, how would I rewrite it with math symbols?

To show that the statement is false, find an $\epsilon$ which does not have a corresponding $x^\star$, such that whenever $x \geq x\star$, $| \frac{1}{x} - 1 | < \epsilon$ does not hold. Suppose $L = 1$ and let $\epsilon = \frac{1}{2}$.
Consider when $x^\star \geq 1$, then
\begin{align*}
| \frac{1}{x} -1 | \geq \frac{1}{2}
\end{align*}
whenever $x \geq 2$. Consider when $x^\star <1$, then
\begin{align*}
|\frac{1}{x} -1 | \geq \frac{1}{2}
\end{align*}
whenever $x \geq 2$.
Hence, there does not exist an $x^\star$ for $\epsilon = \frac{1}{2}$. Therefore, it must certainly be false that the limit is $1$.
