If `lim_(xrarr0) (e^(ax)-e^(x)-x)/(x^(2))=b` (finite), then

Options

A. a = 2, b = 0
B. a = 0, `b=(3)/(2)`
C. `a=2,b=(3)/(2)`
D. a = 0 , b = 2

Answer

C

Explanation

`L=underset(xrarr0)(lim)(e^(ax)-e^(x)-x)/(x^(2))`
`=underset(xrarr0)(lim)(ae^(ax)-e^(x)-1)/(2x)" (Using L'Hospital's Rule)"`
Since denominbator `rarr0`, we must have numerator `rarr0` when `xrarr0`
`therefore" "ae^(0)-e^(0)-1=0 or a=2`
`therefore" "L=underset(xrarr0)(lim)(2e^(2x)-e^(x)-1)/(2x)`
`" "=underset(xrarr0)(lim)(4e^(2x)-1)/(2)" (Using L'Hospital's Rule)"`
`" "=(3)/(2)`