Compute lim x → − 2 ( 3 x 2 + 5 x − 9 )

Answer

`atoq;bto p,q, r;ctor,s;d tor,s.`

Explanation

a. Hence, `agt0`. Therfore.
`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`