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Question-184620




Question Number 184620 by Ml last updated on 09/Jan/23
Commented by Ml last updated on 09/Jan/23
please solution????
pleasesolution????
Answered by mr W last updated on 09/Jan/23
=lim_(x→0) ((sin x^3 )/x^3 )×((tan x)/x)×(((sin (x/2))/(x/2)))^2 ×((x/(sin x)))^5 ×(x^(1/6) /(2(x+x^(1/3) )^(1/2) ))  =lim_(x→0) (x^(1/6) /(2(x+x^(1/3) )^(1/2) ))  =lim_(x→0) (1/(2(x^(4/3) +1)^(1/2) ))  =(1/2)
=limx0sinx3x3×tanxx×(sinx2x2)2×(xsinx)5×x162(x+x13)12=limx0x162(x+x13)12=limx012(x43+1)12=12
Answered by qaz last updated on 10/Jan/23
lim_(x→0^+ ) ((sin x^3 tan x(1−cos x))/( (√(x+(x)^(1/3) ))((x^5 )^(1/6) sin^5 x)))  =lim_(x→0^+ ) ((x^3 ∙x∙(1/2)x^2 )/( ((x)^(1/3) )^(1/2) ∙(x^5 )^(1/6) ∙x^5 ))  =(1/2)  −−−−−−−−−−−  x→0^+ ,sin x^3 ∼x^3     tan x∼x   1−cos x∼(1/2)x^2   (√(x+(x)^(1/3) ))∼((x)^(1/3) )^(1/2)         sin^5 x∼x^5
limx0+sinx3tanx(1cosx)x+x3(x56sin5x)=limx0+x3x12x2x32x56x5=12x0+,sinx3x3tanxx1cosx12x2x+x3x32sin5xx5
Commented by Ml last updated on 09/Jan/23
step by step ????
stepbystep????
Answered by cortano1 last updated on 10/Jan/23

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