l'hopital's rule (1 Viewer)

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,644
Gender
Male
HSC
N/A
Originally posted by KeypadSDM
Lim[x->0] x<sup>x</sup>
= Lim[x->0] e<sup>xLn[x]</sup>

Lim[x->0] xLn[x]
= Lim[x->0] x * Lim[x->0] Ln[x]

D'oh.
Actually, you don't need the rest to do this limit - dominance rules dictate that polynomials dominate logs, so the behaviour of x will dictate the outcome.

Thus, lim (x->0) xln x = 0

You could also note that y = x<sup>x</sup> has a minimum turning point at (1 / e, e<sup>-1/e</sup>), meaning lim (x->O<sup>+</sup>) x<sup>x</sup> > 0, and hence lim (x->O<sup>+</sup>) dy/dx = - inf. So, y = x<sup>x</sup> is descending nearly vertically from the y-axis to a minimum, and then increases.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top