Answer
$L(x)=-x+2$ at $x=x_{0}$
Work Step by Step
$f(x)=\frac{1}{x} $,
$L(x)=f(a)+f'(a)(x-a)$
$f'(x)=-\frac{1}{x^2}$
at $x_{0}=1$ the function and its derivative is close to the value 0.9
$x_{0}=1$
$f(1)=1$
$f'(1)=-1$
then $L(x)=1-1(x-1)=-x+2$
$L(x)=-x+2$ at $x=x_{0}$
Thus, the final answer is: $L(x)=-x+2$ at $x=x_{0}$