Match the following lists:

Answer

`atos;btor;cto p;d toq.`

Explanation

a. `Let" x"+1=h`. Then,
`underset(xto-1)lim(root(3)((7-x))-2)/((x+1))=underset(hto0)lim((8-h)^(1//3)-2)/(h)`
`=underset(hto0)lim(2(1-(h)/(8))^(1//3)-2)/(h)`
`=2underset(hto0)lim((1-(1)/(3)(h)/(8))-1)/(h)`
`=-(1)/(12)`
b. We have
`underset(xto pi//4)lim(tan^(3)x-tanx)/(cos(x+pi//4))`
`=underset(xto pi//4)lim(tanx(tanx-1)(tanx+1))/(cos(x+pi//4))`
`=underset(xto pi//4)lim(tanx(sinx-cosx)(tanx+1))/(cosxcos(x+pi//4))`
`=-underset(xto pi//4)lim(tanx(cosx-sinx)(tanx+1))/(cosxcos(x+pi//4))`
`=-sqrt(2)underset(xto pi//4)lim(tanx((1)/(sqrt(2))cosx-(1)/(sqrt(2))sinx)(tanx+1))/(cosxcos(x+pi//4))`
`=-sqrt(2)underset(xto pi//4)lim(tanx(tanx+1))/(cosx)`
`=-sqrt(2)xx2xxsqrt(2)=-4`
`underset(xto1)lim((2x-3)(sqrt(x-1)))/(2x^(2)+x-3)=underset(xto1)lim((2x-3)(sqrt(x)-1))/((2x+3)(x-1))`
`=underset(xto1)lim((2x-3)(sqrt(x)-1))/((2x+3)(sqrt(x)-1)(sqrt(x)+1))`
`=underset(xto1)lim((2x-3))/((2x+3)(sqrt(x)+1))`
`=(2-3)/((2+3)(sqrt(1)+1))=-1//10`
d. `underset(xtooo)lim(logx^(n)-[x])/([x])=underset(xtooo)lim(logx^(n))/([x])-underset(xtooo)lim([x])/([x])`
`=0-1=-1`