# 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?

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!

• The contrapositive is easier to phrase in my mind – Calvin Khor Oct 10 '18 at 19:11
• In the second definition, it should read $|f(x) - L| > \epsilon$ instead of $|f(x)-L| < \epsilon$. – mz71 Oct 10 '18 at 19:13
• Possible duplicate of How to prove infinite limit is limit does not exist using epsilon and delta – user403337 Oct 10 '18 at 19:31
• Thank you @mzg147! I will edit that – Esther Han Oct 10 '18 at 19:32

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?

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, $$|f(x)-L|<\epsilon$$

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.