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?
How to Prove That $\displaystyle \lim_{t \to 0^+}\; e^{- \frac{1}{t}} = 0$?
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0Yes, this is what I intended; I have fixed the question. – 2012-01-23
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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0No problem. +1 by the way for your "Prove this in general" (and your answer as a whole). – 2012-01-23
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).$
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0Typo extripated. – 2012-01-23
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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0And that is the difference between mathematicians and engineers. – 2012-01-23