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Lim x->0^+ (e^x+x)^(1/x) please use L=(e^x+x)^(1/x) ln(L) so on if you are going to do it that way
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lim (e^x - 1)/x = e^0 -1/0 =1-1/0 = (0/0) x->0 l'hopitals rule (e^x-1)'/x' = e^x lim e^x =1 x->0 \[\lim (e^x - 1)/x = e^0 -1/0 =1-1/0 = (0/0) x->0 l'hopitals rule (e^x-1)'/x' = e^x \lim e^x =1 x->0\]
How did you get the exponent down?
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