LH:

(e^x+x)^1/x
lny=1/xln(e^x+x)
lny=(ln(e^x+x))/x
LH = (e^x+1)/(e^x+x)/1
LH = (e^x)/(e^x+1)
LH = (e^x)/(e^x)
lny=1
e^(lny)=e^1
y=e

Roight? My friend keeps coming up with infinity. Won't agree with my math.

Question

LH:

(e^x+x)^1/x
lny=1/xln(e^x+x)
lny=(ln(e^x+x))/x
LH = (e^x+1)/(e^x+x)/1
LH = (e^x)/(e^x+1)
LH = (e^x)/(e^x)
lny=1
e^(lny)=e^1
y=e

Roight? My friend keeps coming up with infinity. Won't agree with my math.

anonymous · Answer

In case anyone's wondering, LH is used to denote when I've used L'Hopital's rule.

anonymous · Answer

ahhhh, forgot to add that it's a limit as x-> (infinity)

So, the full problem is Lim x-> of (e^x+x)^1/x