How do I prove directly from the $\delta$-$\epsilon$ definitions that if $\lim_{x\to 0} f(x) = \infty$, then $\lim_{x\to 0} f(x)$ does not exist?

Question

The definition of $\lim_{x\to 0} f(x) = \infty$ is:

$\forall N\in\mathbb{R}\;\exists \delta\gt 0\; (0<\lvert x\rvert<\delta\implies f(x)>N)$

The definition of "$\lim_{x\to 0} f(x)$ does not exist" is:

$\forall L\in\mathbb{R}\; \exists \epsilon\gt 0\; \forall \delta\gt 0 \; \exists x\in \mathbb{R}\; (0<\lvert x\rvert<\delta\; and\; \lvert f(x)-L \rvert\gt\epsilon)$

The question makes sense intuitively and geometrically, but how do I prove directly from the definitions that the first definition implies the second?

Thank you!

Answer 1

Suppose you had $\lim_{x\to 0} f(x) = L < \infty$ and $\lim_{x\to 0} f(x) = \infty$. Then from the first, applying it with $\epsilon := 1$, you can find $\delta > 0$ such that $|f(x) - L| < 1$ whenever $0 < |x| < \delta$. Similarly, from the second, you can find $\delta' > 0$ such that $f(x) > L + 1$ whenever $0 < |x| < \delta'$.

Now, can you see from here how to reach a contradiction?

Answer 2

Assuming $$\forall N\in\mathbb{R}\;\exists \delta\gt 0\; (0<\lvert x\rvert<\delta\implies f(x)>N)$$

You want to show that

$$\forall L\in\mathbb{R}\; \exists \epsilon\gt 0\; \forall \delta\gt 0 \; \exists x\in \mathbb{R}\; (0<\lvert x\rvert<\delta\; and\; \lvert f(x)-L \rvert\lt\epsilon)$$ is not true.

Suppose your sequence has a limit $L \in R$

Let $N=L+2\epsilon$ where the $\epsilon$ comes from the definition of limit. For this $N$ you have a $\delta $ such that $$(0<\lvert x\rvert<\delta\implies f(x)>N=L+2\epsilon)$$ That is $|f(x)-L|>2\epsilon$ which contradicts, $|f(x)-L|<\epsilon$

Answer 3

Let $f:D\rightarrow R$ be a finction and $\lim_{x\rightarrow 0} f(x)= \infty$.Now if possible let $\lim_{x\rightarrow 0}f(x)$ exists and equals to $L$, then we have a $\delta>0$ for the positive number $1$ such that whenever $0<|x|<\delta ,x\in D$ then $|f(x)-L|<1$ i.e. for $x\in (-\delta,+\delta)\cap D,x\not =0,$ we have $|f(x)|-|L|≤||f(x)|-|L||≤|f(x)-L|<1$, so that on $\{(-\delta,0)\cup (0,+\delta)\}\cap D$ we show $f$ is bounded by $|L|+1$

But by assumption $\lim_{x\rightarrow 0} f(x)= \infty$ i.e. given any $G>0$ we have $\delta>0$ such that whenever $0<|x|<\delta ,x\in D$ we have $f(x)>G$ i.e. one can choose $\{x_n\} $ from $D$ such that $x_n\rightarrow 0$ but $f(x_n)>n$ i.e. $f$ is unbounded in every deleted nbd of $0$.

Hence $\lim_{x\rightarrow 0} f(x)$ doesn't exist.