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
 A: $\lim_{x→0}f(x)=∞$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 $0.01$
A: 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.
A: The reason limits approaching infinity are considered undefined is that infinity is not a number and therefore the limit is not approaching a defined value.
Another reason is that limits are only defined if the limit from the left-hand side equals the limit from the right-hand side (per the definition of a limit). However, you can only approach infinity from one side.
A: 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. 
