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Answer» `UNDERSET(xtooo)lim("("sqrt("("x^(2)-x+1")")-ax-b")""("sqrt(x^(2)-x+1")")+ax+b")")/("("sqrt("("x^(2)-x+1")")+ax+b")")` `=0` or `underset(xtooo)lim((x^(2)-x+1)-(ax+b)^(2))/(sqrt((x^(2)-x+1))+ax+b)=0` or `underset(xtooo)lim((1-a^(2))x^(2)-(1+2ab)x+(1-b^(2)))/(sqrt((x^(2)-x+1))+ax+b)=0` or `underset(xtooo)lim((1-a^(2))x-(1+2ab)+((1-b^(2)))/(x))/(sqrt(1-(1)/(x)+(1)/(x^(2))+a+(b)/(x)))=0` This is POSSIBLE only when `1-a^(2)=0" and "1+2ab=0` `:.""a=+-1` or `a=1""(becauseagt0)""(1)` `:.""b=-1//2` `:.""(a,2b)-=(1,-1)` b. Divide numerator and denominator by `e^(1//x)`. Then, `underset(xtooo)lim((1+a^(3))e^(-(1)/(x))+8)/(e^(-(1)/(x))+(1-b^(3)))=2` or `(0+8)/(0+1-b^(3))=2` or `""1-b^(3)=4` `:.""b^(3)=-3" or "b=-3^(1//3)` Then, `ainR.` Therefore, `(a,b^(3))-=(a,-3)` c. `underset(xtooo)lim"("sqrt("("x^(4)-x^(2)+1)")"-ax^(2)-b")"=0` Put `s=(1)/(t)`.Then `underset(t TO0)lim(sqrt(((1)/(t^(4))-(1)/(t^(2))+1))-(a)/(t^(2))-b)=0` or `underset(t to0)lim(sqrt((1-t^(2)+t^(4)))-a-bt^(2))/(t^(2))=0""(1)` Since R.H.S. is finite, numerator must be equal to 0 at `t to0.` Therefore,`1-a=0" or "a=1.` From equation (1), `underset(t to0)lim(sqrt((1-t^(2)+t^(4)))-1-bt^(2))/(t^(2))=0` `underset(t to0)lim(-1+t^(2))(((1-t^(2)+t^(4))^(1//2)-(1)^(1//2))/((1-t^(2)+t^(4))-1))=b` or `(-1)((1)/(2))=b" or "a=1,b=-(1)/(2)" or "(a,-4b)-=(1,2)` d. `underset(xto-a)lim(x^(7)-(-a)^(7))/(x-(-a))=7" or "7a^(6)=7" or "a^(6)=1" or "a=-1` |
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