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?
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4$\begingroup$ Your limit is correct if by "$t\to0$" you mean $t\downarrow0$. But $\lim\limits_{t\uparrow0}\ e^{-1/t}\ne0$. $\endgroup$– Michael HardyJan 23, 2012 at 1:54
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$\begingroup$ Yes, this is what I intended; I have fixed the question. $\endgroup$– ItsNotObviousJan 23, 2012 at 2:17
3 Answers
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.
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$\begingroup$ No problem. +1 by the way for your "Prove this in general" (and your answer as a whole). $\endgroup$– JavaManJan 23, 2012 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 $$ 0 < t < 1/\log(1/\epsilon).$$
Make $s=1/t$ such that
$$ \lim_{t \to 0^+}\; e^{- \frac{1}{t}} = \lim_{s \to +\infty}\; e^{-s} = e^{-\infty} =0 $$
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2$\begingroup$ 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!) $\endgroup$ Jan 23, 2012 at 2:55
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$\begingroup$ And that is the difference between mathematicians and engineers. $\endgroup$ Jan 23, 2012 at 14:48