Question #27259

Question

1509 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-20T15:27:14

#e^(-4)#

#Lim_(xrarroo)((3x-4)/(3x+2))^(2x-4)#

#Lim_(xrarroo)((3x+2-6)/(3x+2))^(2x-4)#

#Lim_(xrarroo)((3x+2)/(3x+2)+(-6)/(3x+2))^(2x-4)#

#Lim_(xrarroo)(1+1/((3x+2)/(-6)))^(2x-4)#

#Lim_(xrarroo)(1+1/((3x+2)/(-6)))^((2x-4)(3x+2)/(-6)*(-6)/(3x+2))#

#Lim_(xrarroo)[(1+1/((3x+2)/(-6)))^((3x+2)/(-6))]^((2x-4)*(-6)/(3x+2))#

#Lim_(xrarroo)(1+1/x)^x=e#

#e^(Lim_(xrarroo)((-12x+24)/(3x+2)))#

#e^(-12/3)=e^(-4)#

Answer 2 · 2017-12-20T16:03:40

#e^-4#

Making

#1+1/y = ((3 x - 4)/(3 x + 2))# or equivalently

#x = -2/3(3y+1)# we have

#((3 x - 4)/(3 x + 2))^(2 x - 4) equiv (1+1/y)^(-(4y+16/3))=1/( (1+1/y)^(4y+16/3))#

and then

#lim_(x->oo)((3 x - 4)/(3 x + 2))^(2 x - 4) =1/( lim_(y->oo)(1+1/y)^(4y+16/3)) = 1/e^4 = e^-4#