`lim_(x->0) 1/x [int_(y->a)e^(sin^2t) dt-int_(x+y->a)e^(sin^2t)dt]` is equal to

Options

A. `e^(sin^(2)y)`
B. `sin2ye^(sin^(2)y`
C. `0`
D. none of these

Answer

A

Explanation

`lim_(xto0) 1/x [int_(y)^(a)e^(sin^(2)t)dt+int_(a)^(x+y)e^(sin^(2)t)dt]`
`=lim_(xto0)1/x int_(y)^(x+y)e^(sin^(2)t)dt` (`0/0` form)
Apply L'Hopital rule, we get
`lim_(x to0) (e^(sin^(2)(x+y))(1+(dy)/(dx))-^(sin^(2)y)(dy)/(dx))/1`
`e^(sin^(2)y)[1+(dy)/(dx)-(dy)/(dx)]=e^(sin^(2)y)`