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Answer» `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` |
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