Math Question...Not what you are expecting. L'hopitals rule.

[Doc7]

Heya...

You're going to think I'm a terrible engineer, but I'm drawing a blank on a portion of this homework assignment for finding the limit of a fraction approaching infinity on both the top and the bottom, and googling l'hopitals rule has been as yet fruitless (I know for a fact this is the rule that applies in this case)

(s+1) /
(s+2)(s+3)

lim s->infinity


Derivative with respect to s on both the top and bottom, right? Then take the limit as s approaches infinity again, so it's 0, right?

[bobman0330]

you don't need L'hopital's rule. The numerator is 1st order, the denominator is second order. lim = 0.

EDIT: I think you're a [censored] engineer. /images/graemlins/smile.gif

[Doc7]

psht, i kinda of figured that but i also was a little worried about assuming i was right about infinity < infinity-squared.

[bobman0330]

You can avoid most of the infinities
Multiply both the numerator and the denominator by 1/s. Numerator becomes 1 + 1/s. Denominator becomes as + b + c/s. Limit of the numerator is 1 + 0. Limit of denominator is infinity + b + 0. limit = 0.