Menu Close

lim-x-x-2-x-x-1-4-sin-2-x-x-2-sin-3x-

Tinku Tara 0 Comments
Pin




Question Number 133114 by bemath last updated on 19/Feb/21
lim_(x→∞) (((√(x^2 −x))/x) + (1/4)sin ((2/x)))^(x^2 +sin 3x) ?
limx→∞(x2−xx+14sin(2x))x2+sin3x?
Answered by bobhans last updated on 19/Feb/21
L=lim_(x→∞) (((√(x^2 −x))/x) + (1/4)sin ((2/x)))^(x^2 +sin (3x)) = lim_(x→∞) ((√(1−(1/x)))+(1/4)sin ((2/x)))^(x^2 +sin (3x)) let (1/x) = z , z→0^+ =lim_(z→0^+ ) ((√(1−z)) +(1/4)sin (2z))^((1/z^2 )+sin ((3/z))) ln L = lim_(z→0^+ ) ((1/z^2 )+sin ((3/z)))ln ((√(1−z))+(1/4)sin (2z)) = lim_(z→0^+ ) ((((1+z^2 sin ((3/z)))ln ((√(1−z))+(1/4)sin 2z))/z^2 ) by Sandwich theorem we get that lim_(z→0^+ ) (1+z^2 sin ((3/z)))= 1 therefore the limit reduces to ln L= lim_(z→0^+ ) ((ln ((√(1−z))+(1/4)sin (2z)))/z^2 ) ln L= lim_(z→0^+ ) ((−(1/2)(1−z)^(−(1/2)) +(1/2)cos (2t))/(2t((√(1−z))+(1/4)sin 2z))) = lim_(z→0^+ ) (1/( (√(1−z))+(1/4)sin 2z)). lim_(z→0^+ ) ((−(1−z)^(−1/2) +cos 2z)/(4z)) = 1 .lim_(z→0^+ ) ((−(1/2)(1−z)^(−3/2) −2sin 2z)/4) = −(1/8) ; ln L = −(1/8) then L = e^(−1/8) or L = (1/( ((e ))^(1/(8 )) ))
L=limx→∞(x2−xx+14sin(2x))x2+sin(3x)=limx→∞(1−1x+14sin(2x))x2+sin(3x)let1x=z,z→0+=limz→0+(1−z+14sin(2z))1z2+sin(3z)lnL=limz→0+(1z2+sin(3z))ln(1−z+14sin(2z))=limz→0+((1+z2sin(3z))ln(1−z+14sin2z)z2bySandwichtheoremwegetthatlimz→0+(1+z2sin(3z))=1thereforethelimitreducestolnL=limz→0+ln(1−z+14sin(2z))z2lnL=limz→0+−12(1−z)−12+12cos(2t)2t(1−z+14sin2z)=limz→0+11−z+14sin2z.limz→0+−(1−z)−1/2+cos2z4z=1.limz→0+−12(1−z)−3/2−2sin2z4=−18;lnL=−18thenL=e−1/8orL=1e8

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *