From the definition of linear approximation we know that, for x {\displaystyle x} close to a {\displaystyle a} , we can approximation function value f ( x ) {\displaystyle f(x)} by the following form f ( x ) ≈ L ( x ) = f ( a ) + f ′ ( a ) ( x − a ) . {\displaystyle f(x)\thickapprox L(x)=f(a)+f'(a)(x-a).} Here in this problem we have f ( x ) = ln ( x ) , f ′ ( x ) = 1 x , a = 1 , x = 0.95 , {\displaystyle f(x)=\ln(x),\quad f'(x)={\frac {1}{x}},\quad a=1,\quad x=0.95,} Substitute them into above form we get
L ( 0.95 ) = 0 + 1 ( 0.95 − 1 ) = − 0.05. {\displaystyle L(0.95)=0+1(0.95-1)=-0.05.}
close to 
, we can approximation function value 
by the following form 
Here in this problem we have 
Substitute them into above form we get
