How to answer Calculus by Michael Spivak Chapter 5 probem 17.a

Question

I know this must be easy but I am learning.

Prove that

$$\lim_{x\to 0} \frac{1}{x} $$

Does not exist.

What I did was: Say $$\lim_{x\to 0} \frac{1}{x} = L$$

• ∃ ε > 0 : ∀ δ > 0 ∃x with |x| < δ and |$\frac{1}{x}$ -L| ≥ ε

I do not know how to proceed, how can I find that ε?

Answer 1

Take $\epsilon=1$, for each $\delta>0$ we find an $N$ large enough such that $1/N<\min\{\delta,1/(|L|+1)\}$, then $1/N<1/(|L|+1)$, so $N>|L|+1$ and hence $|1/(1/N)-L|\geq N-|L|>1$.

Answer 2

Look at the differnce between the right limit and left limit as $x\to 0$

One is diverging to $ +\infty $ and the other to $- \infty$