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Evaluate $$\lim_{n\to0}\frac{1}{n}$$

How do I show that the limit does not exist?

I did $\displaystyle \lim_{n\to 0^+}=\infty$ and $\displaystyle \lim_{n\to0^-}=-\infty$ so $\displaystyle \lim_{n\to0^+}\neq \lim_{n\to0^-}$

Thus $\displaystyle \lim_{n\to0}\frac{1}{n}$ does not exist. Am I correct?

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4 Answers

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If you go for extended real line where $\infty $ is included.

Then as, $\lim_{x \to 0+} \frac{1}{x} \to \infty $ and $\lim_{x \to 0-} \frac{1}{x} \to -\infty $ ,

So,from these , you can conclude that the limit doesn't exist even in extended real line $\mathbb{R} ∪ \{\infty\} $

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I agree with the analysis of the other responses. However, I would instead have taken a (very basic) $\epsilon, \delta$ approach of proof by contradiction.

Suppose the limit exists, and let $L$ = this limit.
Without loss of generality, assume that $L \geq 0.$
That is, the proof for $L < 0$ would be very similar.

Then, for all $\epsilon > 0$ there must exist a $\delta > 0$such that $|L - \frac{1}{n}| < \epsilon$ whenever$0 < |(n - 0)| < \delta$.

Choose $\epsilon = (1/2).$
Then supposedly, a satisfying $\delta$ can be found
that corresponds to $\epsilon = (1/2).$

It now remains to show that no matter how small you make $\delta$, it will not satisfy the requirement for $\epsilon = (1/2).$

Choose $M \in \mathbb{Z^+}$ such that both of the following are true:
(1) $M > L + 1$
(2) $(1/M) < \delta.$

Let $N = (1/M).$
Then $0 < |(N - 0)| < \delta$ and
$|L - \frac{1}{N}| = |L - M| > 1 > \epsilon.$

With $\epsilon$ fixed at $\epsilon = (1/2)$
the above analysis shows that no matter how small you choose $\delta,$a positive integer $M$ can be found, and $N$ can be set to $(1/M)$ in such a way as to prove that $\delta$ does not satisfy $\epsilon = (1/2).$

Therefore, with $\epsilon = (1/2)$
no satisfying $\delta$ can be found.

This contradicts the $\epsilon, \delta$ definition of convergence.
Therefore, the limit does not exist.

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Actually for proving not existing a limit it will be enough to make a sequence example that the limits converges for that. For instance you can say: $n = a_m = \frac{1}{m}$. So since $\lim_{m \to \infty} m = \infty$ so the limit does not exist. By the way your provement is correct.

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Your final answer is correct but the way to prove it depends upon the definition you are adopting for the "limit existence".

In some contexts, the definition adopted is that limit at a point exists when right and left limit are equal and its value is a real number. According to this definition, it suffices to observe that

$$\lim_{n\to 0^+}\frac1n=\infty$$

to conclude that limit doesn't exist.

A different definition assumes that limit exists when right and left limit are equal and its value is a real number or it diverges at positive infinite or it diverges at negative infinite. According to this other definition, we need to observe that

$$\lim_{n\to 0^+}\frac1n=\infty \neq \lim_{n\to 0^-}\frac1n=-\infty$$

to conclude that limit at $n=0$ doesn't exist.

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