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Question Number 162672 by LEKOUMA last updated on 31/Dec/21

Calculate  lim_(x→+∞) (((x^3 +3x^2 ))^(1/3) −(√(x^2 −2x)))  lim_(x→7) (((√(x+2))−((x+20))^(1/3) )/( ((x+9))^(1/4) −2))

Calculatelimx+(x3+3x23x22x)limx7x+2x+203x+942

Answered by tounghoungko last updated on 31/Dec/21

(2) lim_(x→7)  (((√(x+2))−3+3−((x+20))^(1/3) )/( ((x+9))^(1/4) −2))  L_1  = lim_(x→7)  ((x−7)/(((√(x+2))+3)(((x+9))^(1/4) −2)))   L_1 = lim_(x→7)  (((x−7)4)/(6((√(x+9))−4))) = (2/3)lim_(x→7)  ((8(x−7))/((x−7)))=((16)/3)    L_2 =lim_(x→7)  ((3−((x+20))^(1/3) )/( ((x+9))^(1/4) −2)) = lim_(x→7)  (((7−x))/(27)).lim_(x→7)  (4/( (√(x+9))−4))  L_2 = (4/(27)). lim_(x→7)  ((7−x)/(x−7)) .(8)=−((32)/(27))   ∴ L=L_1 +L_2 = ((16)/3)−((32)/(27)) = ((112)/(27))

(2)limx7x+23+3x+203x+942L1=limx7x7(x+2+3)(x+942)L1=limx7(x7)46(x+94)=23limx78(x7)(x7)=163L2=limx73x+203x+942=limx7(7x)27.limx74x+94L2=427.limx77xx7.(8)=3227L=L1+L2=1633227=11227

Commented by LEKOUMA last updated on 31/Dec/21

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Answered by tounghoungko last updated on 31/Dec/21

(1) lim_(x→∞)  x ((1+(3/x)))^(1/3) −x (√(1−(2/x)))    [ (1/x)=y ]    L= lim_(y→0)  ((((1+3y))^(1/3) −(√(1−2y)))/y)    L= lim_(y→0)  (((1+((3y)/3))−(1−((2y)/2)))/y) = 2

(1)limxx1+3x3x12x[1x=y]L=limy01+3y312yyL=limy0(1+3y3)(12y2)y=2

Commented by LEKOUMA last updated on 31/Dec/21

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