Step 1:

The function is $\"f(x)=\\ln$ .

The linear approximation of a function is $\"L(x)=f(a)+f\'(a)(x-a)\"$ .

Find the derivative of the function at $\"x=a\"$ .

$\"\\\\$

Now put $\"x=a\"$ in $\"f(x)=\\ln$ .

$\"f(a)=\\ln$

Step 2:

$\"L(x)=f(a)+f\'(a)(x-a)\"$

$\"L(x)=\\ln\\$

Now find the linear approximation at $\"a=0\"$ .

$\"\\\\L(x)=\\ln\\$

So the linear approximation at $\"a=0\"$ is $\"\\\\L(x)=x\"$ .

Step 3:

Now use the linear approximation to find the $\"\\ln$ and $\"\\ln$ .

Rewrite $\"\\ln$ .

Now compare it with the function $\"\\ln$ .

Here $\"x=0.2\"$ .

Now approximate for $\"x=0.2\"$ .

$\"\\\\L(x)=x$

Step 4:

Rewrite $\"\\ln$ .

Now compare it with the function $\"\\ln$ .

Here $\"x=0.01\"$ .

Now approximate for $\"x=0.01\"$ .

$\"\\\\L(x)=x$

Step 5:

Now calculate $\"\\ln$ and $\"\\ln$ using calculator.

$\"\\ln$

$\"\\ln$

The second approximation is significantly better, since it is significantly closer to $\"0\"$ than $\"1.2\"$ .

Solution :

(1) The linear approximation is $\"\\\\L(x)=x\"$ .

(2) $\"\\\\L(0.2)=0.2\"$ .

(3) $\"\\\\L(0.01)=0.01\"$ .

(4) The second approximation is more accurate.