Question

How to calculate lim e^(-1/(t^2))/t t->0 ?

Answer 1

\[\Large \lim_{t\to0}{e^{-1\over t^2}\over t}\]
substitute \(u={1\over t}\)
as \(t\to0,u\to\pm\infty\)

Answer 2

thanks

Answer 3

hmm, I would say the limit is 0.
\[\lim_{t \rightarrow 0} e ^{-\frac{ 1 }{ t^2 }} = 0\]
And this reach '0' much faster than 't' (the denominator), so 0 is reached in the numerator 'first'.
Since you have 0/0 I tried using l'hopital, but I cant get it into a form that proves this.

Remember lim x -> 0, means really really close to zero, but not including zero, so yes, it is undefined at 0, but really really close to zero the value is 0.
So the limit as x approaches 0 is 0 (in my opinion)

Answer 4

Hi thank you jkristia
I think I found a way for salving with change variable
as electrokid said u=1/t
then \[\lim_{t \rightarrow 0}=\lim_{u \rightarrow \infty}=\lim_{u \rightarrow \infty}e^{-u^{2}}/1/u=\lim_{u \rightarrow \infty}u/e^{u^{2}} -->Hopital =\lim_{u \rightarrow \infty}1/2ue^{u^{2}}=1/\infty=0\]

Answer 5

yes, now when I see it, that looks right, and shows the limit is 0