$\lim_{t\rightarrow \infty} e^{-lt}$ limit

We know that if $K<0$, then there exists $l>0$, such that $K<-l$. What is $\displaystyle\lim_{t\rightarrow \infty} e^{-lt}$? It seems to be $0$, but I'm not sure because $l$ can be $<<$, then we will face to the case of indefinite limit. Clear me please about this point.

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And, what is $l$? Can it be regarded fixed and $l\ne 0$? If not, then yes, what's written could also allow $l\to 0$ and then $\lim_{t\to\infty}lt$ can be anything. – Berci Feb 27 '13 at 11:24

Assuming $l$ is a constant, then $\displaystyle \lim_{t\rightarrow \infty}e^{-lt}$=0. This is because $t$ is tending to infinity whereas $l$ is a fixed number. It is irrelevant how small $l$ is, as we can always find larger numbers. (Thus $lt$ tends to infinity itself.)
You said yourself: $l > 0$ – Stefan Feb 27 '13 at 15:38