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

Jul 31, 2015

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

#### Explanation:

The linear approximation of $f \left(x\right)$ at $x = a$ is given by:

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

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

In point-slope form, the line is:

$y - f \left(a\right) = f ' \left(a\right) \left(x - a\right)$

So we can write the tangent line as: $y = f \left(a\right) + f ' \left(a\right) \left(x - a\right)$

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 $f \left(x\right) = \ln x$, we have $f ' \left(x\right) = \frac{1}{x}$.

Therefore, $f ' \left(1\right) = \frac{1}{1} = 1$

We also not that $f \left(1\right) = \ln \left(1\right) = 0$.

The linear approximation is the line:

$y - 0 = 1 \left(x - 1\right)$

Or, simply $y = x - 1$

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

$\left.\begin{matrix}x & \text{ calculator" ln(x) & " approx by "x-1 \\ 1.05 & " "" "0.04879 & " "" "0.05 \\ 1.01 & " "" "0.00995 & " "" "0.01 \\ 0.997 & " "-.0.003005 & " } - 0.003\end{matrix}\right.$

Note, however that $\ln \left(2\right) \approx 0.6931$ While the linear approximation gives $2 - 1 = 1$ (not a very good approximation for many purposes)