limit of (e^x+x)^(1/x) as x approaches 0

I remember using e^L where L=(e^x+x)^(1/x) for this 1^infinity form

what do I do now?

[math]\lim_{x{\to}0}(e^x+x)^{\frac{1}{x}}[/math] is much better than that mess. well, e⁰=1, so you get 1^1/0. i would say the answer is 1.

heh, I just found out what I forgot, I forgot to balance out the e by adding natural log. so it would be e^L=e^(ln(e^x+x)*(1/x)) and then by using properties of log and change L to quotient to get (e^x+x)/x so that it is form infinity/infinity and now using L Hopital's rule to get (1+1)/1 so it is 2 but 2=L so the answer would be e^2. woot...

Let consider:

Lim x -> 0 {f(x)}^g(x), where as x -> 0 ,f(x) -> 1,and g(x) ->infinity

lim x->0 [f(x)]^g(x)=lim x->0[1+f(x)-1]^{{g(x)(f(x)-1)}/(f(x)-1)}

= e^lim x ->0{g(x)(f(x)-1)}

here ,we consider only the part; lim x->0 (e^x)^(1/x)

f(x)=e^x and g(x)=1/x so limit becomes :

e^[lim x->0 ((e^x-1)/x)=e^1=e :as (e^x-1)/x=1