# Help seeing algebra in calculus

I'm going through an examples section (on improper integrals) but I got lost at this bit:

$$\lim_{t\to-\infty} \frac{t}{e^{-t}} = \lim_{t\to-\infty}\frac{1}{-e^{-t}}.$$

I think it's a simple algebra trick but I don't see how the right hand side came to be. How did it become $1/-e^{-t}$?

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You could verify by applying ten times the same rule (see the answers) that $\dfrac{e^x}{x^{10}}$ tends to $\infty$ with $x$. –  Américo Tavares Nov 15 '10 at 8:54

It is $t\to-\infty$, not $t\to\infty$. If it were $t\to\infty$, it would not be an indeterminate form. –  Jonas Meyer Nov 15 '10 at 5:04