2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Gerry Myerson Oct 10 '18 at 3:09
  • $\begingroup$ Related math.stackexchange.com/questions/2945237 $\endgroup$ – Nosrati Oct 10 '18 at 3:16
1
$\begingroup$

$\lim_{x→0}f(x)=∞$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 $0.01$

$\endgroup$
  • $\begingroup$ but how would that make the limit undefined then? $\endgroup$ – Matt Oct 10 '18 at 3:17
  • $\begingroup$ 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. $\endgroup$ – Mohammad Riazi-Kermani Oct 10 '18 at 3:24
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.