# How to prove infinite limit is limit does not exist using epsilon and delta

So I was recently taught that If $$\lim_{x→0}f(x)=∞$$, then the limit does not exist, can anyone explain that using epsilon and delta if its possible? But honestly any sort of explanation would be fine

• If the limit exists, call it $L$, with $L\ne\infty$. Then the limit is $L$, so it can't be $\infty$. Epsilon and delta not relevant. – Gerry Myerson Oct 10 '18 at 3:09
• – Nosrati Oct 10 '18 at 3:16

$$\lim_{x→0}f(x)=∞$$is defined as

For every $$M>0$$ there exits a $$\delta >0$$ such that if $$0<|x|<\delta$$ then $$f(x)>M$$

That simply means we can make $$f(x)$$ as large as we wish but the price to pay is to make |x| small enough.

For example we can make $$\frac {1}{x^2}$$ larger than $$10000$$ provided that we make $$|x|$$ less than $$0.01$$

• but how would that make the limit undefined then? – Matt Oct 10 '18 at 3:17
• It means that $\infty$ it is not a real number. If the right limit and left limit are different then the limit does not exist or it is undefined, but if both right and left limits are $\infty$ or both are $-\infty$ we better say that the limit is not real instead of saying it is undifined. – Mohammad Riazi-Kermani Oct 10 '18 at 3:24

if the limit exists, say $$\lim_{x \rightarrow 0} f(x) = L$$. then for every $$\varepsilon > 0$$, there should be a $$\delta > 0$$ s.t. when $$|x| < \delta$$, $$|f(x)-L| < \varepsilon$$.

now $$\lim_{x \rightarrow 0}f(x) = +\infty$$ then there exists a $$\delta^\prime > 0$$ s.t. when $$|x| < \delta^\prime$$, $$f(x) > |L|+\varepsilon+1$$. then you see that the above $$\delta$$ does not exist.

If $$\lim_{x\to0}f(x)=\infty$$, then for any $$L$$, and any $$\epsilon\gt0$$, $$\not\exists\delta \gt0$$ such that $$\mid x\mid\lt\delta \implies \mid f(x)-L\mid\lt\epsilon$$. This is because $$\exists x$$ such that $$\mid x\mid\lt\delta$$ and $$f(x)\gt L+\epsilon$$, by definition.