How to Prove That $\displaystyle \lim_{t \to 0^+}\; e^{- \frac{1}{t}} = 0$? 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?
 A: 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.
A: 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
$$ 0 < t < 1/\log(1/\epsilon).$$
A: Make $s=1/t$ such that 
$$ \lim_{t \to 0^+}\; e^{- \frac{1}{t}} = \lim_{s \to +\infty}\; e^{-s} = e^{-\infty} =0 $$
