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Question-138959

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Question Number 138959 by mathlove last updated on 20/Apr/21
Answered by mathmax by abdo last updated on 20/Apr/21
a^(x^2 −1) =e^((x^2 −1)loga) ∼1+(x^2 −1)loga and b^(x^3 −1) =e^((x^3 −1)logb) ∼1+(x^3 −1)logb c^(x^5 −1) ∼e^((x^5 −1)logc) ∼1+(x^5 −1)logc ⇒ f(x)=((a^(x^2 −1) −b^(x^3 −1) )/(c^(x^5 −1) −1))∼((1+(x^2 −1)loga−1−(x^3 −1)logb)/((x^5 −1)logc)) ∼(((x+1)loga−(x^2 +x+1)logb)/((1+x+x^2 +x^3 +x^4 )logc)) ⇒ lim_(x→1) f(x) =((2loga−3logb)/(5logc))
ax2−1=e(x2−1)loga∼1+(x2−1)logaandbx3−1=e(x3−1)logb∼1+(x3−1)logbcx5−1∼e(x5−1)logc∼1+(x5−1)logc⇒f(x)=ax2−1−bx3−1cx5−1−1∼1+(x2−1)loga−1−(x3−1)logb(x5−1)logc∼(x+1)loga−(x2+x+1)logb(1+x+x2+x3+x4)logc⇒limx→1f(x)=2loga−3logb5logc
Answered by mitica last updated on 21/Apr/21
lim_(x→) ((((a^(x^2 −1) −1)/(x^2 −1))(x−1)(x+1)−((b^(x^3 −1) −1)/(x^3 −1))(x−1)(x^2 +x+1))/(((c^(x^5 −1) −1)/(x^5 −1))(x−1)(x^4 +x^3 +x^2 +x+1)))= =((2lna−3lnb)/(5lnc))
limx→ax2−1−1x2−1(x−1)(x+1)−bx3−1−1x3−1(x−1)(x2+x+1)cx5−1−1x5−1(x−1)(x4+x3+x2+x+1)==2lna−3lnb5lnc

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