Question

Calculati. Limita(cand x ->0) din (1+ radical din x ×lnx)^1/x

Answer 1

L = lim x->0 (1+√x•lnx)^(1/x) =

lim x-> 0 √x•ln x =

√x = t => x= t²

lim t->0 2tlnt = lim t->0 (2lnt)/(1/t) = (L'h) = (2/t)/(-1/t²) = (-2t²)/t = 0

L = lim x -> 0 (1+√x•lnx)^[1/(√x•lnx)•(√x•lnx /x)]

= e^lim x-> 0 (√x•lnx / x)

lim x-> 0 (√x•lnx / x) =

= lim x-> 0 (lnx / √x) =

√x = t => x = t² => t -> 0

= lim t -> 0 (ln(t²)/t) =

= lim t -> 0 (2lnt / t) =

ln t = y => t = e^y => y -> -∞

= lim y -> -∞ (2y/e^y) =

= lim y -> -∞ (2y/e^y) = (-∞/0+) = -∞•(1/0+) = -∞•∞ = -∞

L = e^(-∞) = 0