大一微积分数学【极限?

求过程，谢谢

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1、
lim<x→∞>[3x-√(ax^2+bx+1)]=2
===> lim<x→∞>盯散[3x-√(ax^2+bx+1)]*[3x+√(ax^2+bx+1)]/[3x+√(ax^2+bx+1)]=2
===> lim<x→∞>[9x^2-(ax^2+bx+1)]/[3x+√(ax^2+bx+1)]=2
===> lim<x→∞>[(9-a)x^2-bx-1]/[3x+√(ax^2+bx+1)]=2
所以，9-a=0
则，a=9
===> lim<x→∞>(-bx-1)/凯前[3x+√(9x^2+bx+1)]=2
===> lim<x→∞>[(-b)-(1/x)]/[3+√(9+b/x+1/x^2)]=2
===> (-b-0)/[3+√(9+0+0)]=2
===> -b/(3+3)=2
===> b=-12

2、
lim<x→盯则清π>lnsin(x-π/2)
=lnsin(π/2)
=ln1
=0

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3x-根号仔弯态（ax^2+bx+1)= [ 3x-根号（ax^2+bx+1)][3x+根号（ax^2+bx+1)]/[3x+根号（ax^2+bx+1)]=9x^2-ax^2-bx-1/念源[3x+根号（闹唯ax^2+bx+1)] a=9 b=-12
sin(π-π/2）=sinπ/2=1 ln1=0

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(1)a=9,b=-12
(2)0

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