If `f(x)=(sin(e^(x-2)-1))/(1n(x-1))`, then `lim_(xto2)f(x)` is equal to

Options

A. `-2`
B. `-1`
C. 0
D. 1

Answer

D

Explanation

`f(x)=sin(e^(x-2)-1)/(In(x-1))`
`underset(xto2)lim(sin(e^(x-2)-1))/(In(x-))=L`
It is `(0)/(0)` (undefined) condition so using L' hospital's rule
`impliesL=underset(xto2)lim[({sin(e^(x-2)-1)}^(-)]/({In(x-)}^(-))]`
`impliesL=underset(xto2)lim(cos(e^(x-2)-1).e^((x-2)))/(1//(x-1))`
`impliesL=underset(xto2)limcos(e^(2-2)-1)e^(2-2).(2-1)`
`impliesL=cos(0)e^(0).1`
`impliesL=1`