If `L=underset(xtooo)lim(x+1-sqrt(ax^(2)+x+3))` exists finitely then
The value of a is

Answer

B

Explanation

`L=underset(xtooo)lim(x+1-sqrt(ax^(2)+x+3))`
`=underset(xtooo)lim(((x+1)^(2)-(ax^(2)+x+3))/(x+1+sqrt(ax^(2)+x+3)))`
`=underset(xtooo)lim(((1-a)x^(2)+x-2)/(x+1+sqrt(ax^(2)+x+3)))`
L exists finitely if `1-a=0" or "a=1`
`:." "L=underset(xtooo)lim((x-2)/(x+1+sqrt(x^(2)+x+3)))`
`=underset(xtooo)lim((1-(2)/(x))/(1+(1)/(x)+sqrt(1+(1)/(x)+(3)/(x^(2)))))`