If the function `f(x)` defined as `f(x)` defined as `f(x)={3,x=0(1+(a x+b x^3)/(x^2)),x >0` is continuous at `x=0,` then `a=0` b. `b=e^3` c. `a=1` d. `b=(log)_e3`

Options

A. a = 0
B. `b=e^(3)`
C. `a=1`
D. `b=log_(e)3`

Answer

A, D

Explanation

`underset(hrarr0)(lim)f(0+h)=underset(hrarr0)(lim)(1+(ah+bh^(3))/(h^(2)))^(1//h)`
`" "=underset(hrarr0)(lim)e^((1)/(h)ln(1+(ah+bh^(3))/(h^(2))))`
For limit to exist, we must have
`underset(hrarr0)(lim)(ah+bh^(3))/(h^(2))=0`
`rArr" "underset(hrarr0)(lim)(a+bh^(2))/(h)=0`
`therefore" "a=0`
So, we have
`underset(hrarr0)(lim)f(0+h)=underset(hrarr0)(lim)(1+bh)^(1//h)`
`" "=underset(hrarr0)(lim)(1+bh)^((1//bh)b)=e^(b)`
For f(x) to be continuous at x = 0, we must have
`underset(xrarr0^(+))(lim)f(x)=f(0)`
`rArr" "e^(b)=3`
`therefore" "b=log_(e)3`