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This question already has an answer here:

Prove, using the $\epsilon-\delta$ definition of the limit, that $\displaystyle\lim_{x\to 0}\frac{1}{x}$ fails to exist.

It is obvious that $\displaystyle\frac{1}{x}$ approaches $+\infty$ from the right of $0$ and approaches $-\infty$ from the left of $0$. But how do I show that the limit does not exist using the $\epsilon-\delta$ definition?

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marked as duplicate by Ethan Bolker, Strants, Jack's wasted life, Daniel W. Farlow, Omnomnomnom Nov 1 '16 at 20:20

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  • $\begingroup$ Suppose there is a limit, and show that you reach a contradiction. $\endgroup$ – Astyx Oct 30 '16 at 16:16
  • $\begingroup$ You could write the definition of the limit as a logical statement, negate it, and prove the negation. $\endgroup$ – David K Oct 30 '16 at 16:35
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Let $f(x)=\frac{1}{x}$.

Take the sequences

$u_n=\frac{1}{n}$ and $v_n=\frac{-1}{n}$

$(u_n)$ and $(v_n)$ go to $0$.

thus

$\forall \alpha>0\;\; \exists N\in \mathbb N\;\; : \forall n>N $

$-\alpha<v_n<0<u_n<\alpha$.

or

$\forall\alpha>0 \;\;\exists N \in \mathbb N\;\; : \forall n>N$

$f(u_n)>\frac{1}{\alpha}$ and $f(v_n)<\frac{-1}{\alpha}$

For large enough $n$, $f(u_n)$ and $f(v_n)$ couldn't be in the same neighborhood of any element of $\mathbb R \cup\{-\infty,\infty\}$.

thus, if we take $\epsilon=1$ or $A=1$ for example, and $\alpha=1$, there is a contradiction.

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Assume that $\lim_{x \to 0} \frac 1x = L < \infty$.

Now take $\epsilon = 1$. Then we have that there exists $\delta>0$ such that: $$0 < |x-0| < \delta \implies |f(x) - L| < 1$$

Now if $\boxed{L > -1}$ then we take $x_1 = \min \{ \frac{1}{L+1},\delta - \epsilon_1\}$, where $\epsilon_1$ is some small positive value, s.t. $\delta - \epsilon_1 > 0$.

Now obviously $0 < |x_1 - 0| < \delta$, so therefore we have that $|f(x_1) - L| < 1 \implies \frac 1{L+1} < x_1$, but this contradicts the fact that $x_1 \le \frac{1}{L+1}$

Similar reasoning can be applied to $\boxed{L \le -1}$, by taking $x_1 = \min \{ -\frac{1}{L-1},\delta - \epsilon_1\}$, as then $|f(x_1) - L| < 1 \implies f(x_1) - L > -1 \implies f(x_1) > L-1 \implies x_1 < \frac{1}{L-1} < 0$, which is an obvious contradiction.

This is enough to conclude that the limit doesn't exist.

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