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 $1/-e^{-t}$?

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

It is an application of l'Hôpital's rule.

  • $\begingroup$ Doh! Thank you. I totally forgot about taking the derivatives of f(x)/g(x). $\endgroup$ – ShrimpCrackers Nov 15 '10 at 4:47

The ratio of functions which you provide are an example of an indeterminate form at the limit point. L'Hospital's rule allows you to differentiate the numerator and denominator independently and then take the limit, giving the intended limit. For your example, \begin{eqnarray} \lim_{t \to -\infty} \frac{t}{e^{-t}} \stackrel{L.'H.}{=} \lim_{t \to -\infty} \frac{1}{-e^{-t}} = 0. \end{eqnarray}

  • 5
    $\begingroup$ It is $t\to-\infty$, not $t\to\infty$. If it were $t\to\infty$, it would not be an indeterminate form. $\endgroup$ – Jonas Meyer Nov 15 '10 at 5:04
  • $\begingroup$ Thanks. I corrected the typo. $\endgroup$ – user02138 Nov 15 '10 at 17:20

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