Clarify books proof limit of $\frac{1}{x}$ diverging at $0$

Question

My book shows that the limit diverges at $0$ obviously. How they do it, though, is a bit baffling to me.

Define $f:(0,1)\to \mathbb{R}$ by $f(x)=\frac{1}{x}$. $0$ is an accumulation point so we may inquire as to the existence of the limit there. Let $L$ by the limit, and choose $\epsilon>0$ such that $L+\epsilon>0$. Now, if $0<x<\frac{1}{L+\epsilon}$, then $L+\epsilon<\frac{1}{x}$; hence $|f(x)-L|<\epsilon$. Thus it is impossible to find a $\delta >0$ to fulfill the requirement of the definition.

So I understand if $0<x<\frac{1}{L+\epsilon}$ is true then we have it but how can we guarantee this is the case? I need some further clarification of the details. Is there another way beginning from $0<|x|<\delta$ implies $\left| \frac{1}{x}-L\right| < \epsilon$? Thanks for your help!

Answer 1

What does it mean for $f(x) = 1/x$ to have a limit $L$ at $x = 0$? For $\lim \limits_{x \to 0} f(x) = L$, we need

$$\text{for every }\epsilon > 0\text{, } \exists \delta > 0 \text{ such that for all } x\text{ with } |x - 0| < \delta\text{, } |f(x) - L| < \epsilon.$$

So, to prove that $f$ does not have a limit as $x \to 0$, let's prove that for each real number $L$, the above definition fails (so that for each real number $L$, $L$ is not the limit of $1/x$ as $x \to 0$).

Ok, so let $L$ be an arbitrary real number. It could be positive, negative, or $0$. We need to show that the above definition is false, i.e., the negation is true. But the negation is:

$$\exists \epsilon > 0 \text{ such that } \forall \delta > 0 \text{, } \exists x \text{ such that } |x - 0| < \delta \text{ and } |f(x) - L| \geq \epsilon $$

Ok, so we need to find some $\epsilon > 0$ such that the above is true. Well, let $\epsilon$ be any positive real number such that $L + \epsilon > 0$. (Remember, $L$ could be positive, negative, or $0$.)

Then I need to show for every $\delta > 0$, I can find $x$ such that $x$ is less than $\delta$ from $0$ (i.e., $|x - 0| < \delta$) and $f(x)$ is at least $\epsilon$ from $L$ (i.e., $|f(x) - L| \geq \epsilon$).

Now, we can assume without loss of generality that $\delta < \dfrac{1}{L + \epsilon}$ (why???). That means we only have to find $x$ for each $0< \delta < \dfrac{1}{L + \epsilon}$.

Then, if $|x - 0| < \delta$, we have $|x| < \dfrac{1}{L + \epsilon}$, i.e., $L + \epsilon < \frac{1}{|x|} = \left | \dfrac{1}{x} \right |$, i.e., $\epsilon < \left | \dfrac{1}{x} \right | - L = \left | \dfrac{1}{x} - L \right |$ (why does the last equality hold??).

So, for all of the $\delta < \dfrac{1}{L + \epsilon}$, we didn't just find one $x$ such that $|x - 0| < \delta$ and $|f(x) - L| \geq \epsilon$. We showed that for every $x$ satisfying $|x - 0| < \delta$, $|f(x) - L| \geq \epsilon$ (which is more than we needed to show -- we just needed to find one for each $\delta$). In any case, we showed what we needed to, and so for any real number $L$, it can't be a limit of $\dfrac{1}{x}$ as $x \to 0$.