Using $\varepsilon - \delta$, how would I show $\lim\limits_{x \to 0}f(x)$ has no limit?

Question

$\text{Does }\lim\limits_{x \to 0}\frac{7}{x}\text{ exist?}$

I know that the limit doesn't exist, but don’t know how to prove it. I don’t really understand how to use the $\varepsilon -\delta$ definition of a limit to solve this. Any pointers would be helpful.

Answer 1

For any $\;\epsilon>0\;$ there exists $\;\delta=\dfrac{7}{\epsilon}>0\;$ such that

$0<|x|<\delta\implies\big|f(x)\big|=\left|\dfrac{7}{x}\right|>\dfrac{7}{\delta}=7\dfrac{\epsilon}{7}=\epsilon\;.$

Hence there exists $\;\lim\limits_{x\to0} \big|f(x)\big|=\lim\limits_{x\to0} \left|\dfrac{7}{x}\right|=+\infty\;.$

Analogously we can prove that there exist

$\lim\limits_{x\to0^+} f(x)=\lim\limits_{x\to0^+}\dfrac{7}{x}=+\infty\;,\quad\color{blue}{(*)}$

$\lim\limits_{x\to0^-} f(x)=\lim\limits_{x\to0^-}\dfrac{7}{x}=-\infty\;.\quad\color{blue}{(**)}$

Now we are going to prove the limit $\;(*)\;$.

For any $\;\epsilon>0\;$ there exists $\;\delta=\dfrac{7}{\epsilon}>0\;$ such that

$0<x<\delta\implies f(x)=\dfrac{7}{x}>\dfrac{7}{\delta}=7\dfrac{\epsilon}{7}=\epsilon\;.$

Now we are going to prove the limit $\;(**)\;$.

For any $\;\epsilon>0\;$ there exists $\;\delta=\dfrac{7}{\epsilon}>0\;$ such that

$-\delta<x<0\implies f(x)=\dfrac{7}{x}<-\dfrac{7}{\delta}=-7\dfrac{\epsilon}{7}=-\epsilon\;.$

We have just proved the limits $\;(*)\;$ and $\;(**)\;.$

Since the results of the limits $\;(*)\;$ and $\;(**)\;$ are different $(+\infty$ and $-\infty)$, there does not exist the limit $\;\lim\limits_{x\to0}f(x)=\lim\limits_{x\to0}\dfrac{7}{x}\;.$

Answer 2

If this limit exists, let us call him $L$. By definition, $\forall\varepsilon>0$ there exist $\delta>0$ such that, $x\neq0$ and $|x-0|=|x|<\delta\implies|\frac{7}{x}-L|<\varepsilon$. Taking, for example, $\varepsilon=1$, there exist $\delta=\delta(1)$ such that $$x\neq0, |x|<\delta\implies|\frac{7}{x}-L|<1(*).$$ On the other hand, Let $n$ be the least positive integer $m\in\mathbb{N}$ such that $\frac{1}{m}<\delta$. Observe that we obtain the increasing sequence $$f(\frac{1}{n})=7n<f(\frac{1}{n+1})=7(n+1)<f(\frac{1}{n+2})=7(n+2)<\cdots$$ which clearly will overcome $L+1$. But, for all $m\geqslant n$, we have $\frac{1}{m}\leqslant\frac{1}{n}<\delta$, and we get a contradiction for $(*)$.