wrong limit



On Wednesday 31 December 2008 16:11:04 Jaime Villate wrote:

> By the way, this example puzzles me because the function is clearly
> negative for x>7, so the limit must be minf, but it can also be
> written as:
> limit((7-x)/exp(4-x),x,inf)
> and applying L'Hopytal's rule:
> limit(1/exp(4-x),x,inf)
> which is inf. What's wrong with this?

l'Hopital's rule could be used for \limit f(x)/g(x) where

  \limit f(x) = \limit g(x) = 0

or

  \limit f(x) = \limit g(x) = \pm \infty

But for (7-x)/exp(4-x) you have -\infty and 0.

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Setting Orange, Aftermath 73 YOLD 3174
Alexey Beshenov  http://beshenov.ru/