Question

How do you find the linear approximation of `f(x)=ln(x)` at `x=1` ?

Answer 1

Here is the big key: The linear approximation of `f` at `a` is the tangent line at `a`.

Explanation:

The linear approximation of `f(x)` at `x=a` is given by:

`L(x) = f(a) + f'(a) (x-a)`

The equation of the tangent line to the graph of `f` at `(a, f(a))` is the equation of a line through `(a, f(a))` whose slope is `f'(a)` .

In point-slope form, the line is:

`y-f(a) = f'(a)(x-a)`

So we can write the tangent line as: `y = f(a) +f'(a)(x-a)`

I struggled mightily with linear approximation for a week or two before this hit home and I realized that linear approximation is just a particular way of thinking about, writing, and using the tangent line. (My teacher had been saying it, I just didn't see it.)

In this problem

For `f(x) = lnx`, we have `f'(x) = 1/x`.

Therefore, `f'(1) = 1/1 = 1`

We also not that `f(1) = ln(1) =0`.

The linear approximation is the line:

`y-0 = 1(x-1)`

Or, simply `y = x-1`

If you have a calculator of tables for `ln` you can quickly see that

`{: (x," calculator" ln(x), " approx by "x-1),(1.05," "" "0.04879," "" "0.05),(1.01," "" "0.00995," "" "0.01),(0.997," "-.0.003005," "-0.003) :}`

Note, however that `ln(2) ~~ 0.6931` While the linear approximation gives `2-1 = 1` (not a very good approximation for many purposes)