Tell me more ×

What is the best way to prove that $\displaystyle \lim_{t \to 0^+}\; e^{- \frac{1}{t}} = 0$? Intuitively, it seems true because as $t \rightarrow 0$ from above, $\frac{1}{t} \rightarrow \infty$ and therefore $e^{-\frac{1}{t}} \rightarrow 0$. Is there a way to turn these observations into a rigorous proof without pulling some magic epsilon from the aether?

share|improve this question
Your limit is correct if by "$t\to0$" you mean $t\downarrow0$. But $\lim\limits_{t\uparrow0}\ e^{-1/t}\ne0$. – Michael Hardy Jan 23 '12 at 1:54
Yes, this is what I intended; I have fixed the question. – ItsNotObvious Jan 23 '12 at 2:17

3 Answers

up vote 3 down vote accepted

Assuming we take into account Michael's observation in the comment above, what you wrote is a proof, provided you know how to justify the deductions involved.

Indeed, it is true that

if $\lim\limits_{t\to b}f(t)=a$ and $\lim\limits_{s\to c}g(s)=b$, then $\lim\limits_{s\to c}f(g(s))=a$.

Prove this in general. Moreover, this is also true when some of $a$, $b$ and $c$ are not numbers but $+\infty$ or $-\infty$, and when some of the limits have the arrow $\to$ replaced by $\uparrow$ or $\downarrow$, provided you combine things correctly. (It is probably a useful excercise to make the complete list of statements of this form that are true, in fact!)

Once you have that, then prove that $\lim_{t\downarrow0}1/t=+\infty$ and that $\lim_{t\to+\infty}e^{-t}=0$.

Finally, put things together.

share|improve this answer
Thanks! ${}{}{}$ – Mariano Suárez-Alvarez Jan 23 '12 at 3:35
No problem. +1 by the way for your "Prove this in general" (and your answer as a whole). – JavaMan Jan 23 '12 at 3:37

To prove that $\lim_{t\downarrow 0} e^{-1/t}$ = 0, set $$e^{-1/t} < \epsilon.$$ Taking reciprocals, $$e^{1/t} > 1/\epsilon.$$ Now take logs on both sides to get $$1/t > \log(1/\epsilon).$$ Finally invert to get $$ 0 < t < 1/\log(1/\epsilon).$$

share|improve this answer
Typo extripated. – ncmathsadist Jan 23 '12 at 2:24

Make $s=1/t$ such that

$$ \lim_{t \to 0^+}\; e^{- \frac{1}{t}} = \lim_{s \to +\infty}\; e^{-s} = e^{-\infty} =0 $$

share|improve this answer
Using $e^{-\infty}$ is nothing more that a notation which only serves to hide one of the arguments in the other answers... It should never appear in a «rigorous proof», which is what the OP wants (at least, at the level of someone asking this type of questions!) – Mariano Suárez-Alvarez Jan 23 '12 at 2:55
And that is the difference between mathematicians and engineers. – ja72 Jan 23 '12 at 14:48

Your Answer


