Question Number 103463 by bemath last updated on 15/Jul/20

lim_(x→∞)  ((ln(3e^(2x) +5e^x −2))/(ln(27e^(3x) −1))) ?

Answered by Worm_Tail last updated on 15/Jul/20

lim_(x→∞)  ((ln(3e^(2x) +5e^x −2))/(ln(27e^(3x) −1))) =lim(((6e^(2x) +5e^x )/(3e^(2x) +5e^x −2))/((54e^(3x) )/(27e^(3x) −1)))  lim((6+(5/e^x ))/(3+(5/e^x )−(2/e^(2x) )))×((27−(1/e^(3x) ))/(54))=(6/3)×((27)/(54))=(2/3)

Answered by bobhans last updated on 15/Jul/20

L=lim_(x→∞)  ((ln(e^(2x) (3+5e^(−x) −2e^(−2x) )))/(ln(e^(3x) (27−e^(−3x) ))))  L=lim_(x→∞)  ((ln(e^(2x) )+ln(5e^(−x) −2e^(−2x) ))/(ln(e^(3x) )+ln(27−e^(−3x) )))  L= lim_(x→∞)  ((2x+ln(3+5e^(−x) −2e^(−2x) ))/(3x+ln(27−e^(−3x) )))  L= lim_(x→∞) ((2+(1/x)ln(3+5e^(−x) −2e^(−2x) ))/(3+(1/x)ln(27−e^(−3x) )))  note lim_(x→∞)  (1/e^x ) = lim_(x→∞) e^(−x)  = 0  L=((2+0.ln(3+0−0))/(3+0.ln(27−0))) = (2/3). (⊕)