Linear approximation $y= \ln(1+x)$ for small x

Question

How can I show with linear approximation that $y \approx x$ for small x? I know the rule $$f(x) \approx f(a) + f^{\prime}(a) (x-a),$$ but I don't know how to put it to use in this case.

Answer 1

The function you are trying to approximate is $$ f(x) = \ln(x) $$ and you need an approximation around $a = 1$.

For the rule we need the derivative of the function, and we know that the derivative of the natural logarithm is the inverse: $$ f'(x) = \frac1x. $$

Let's evaluate both the function and the derivative at $a = 1$, $$ f(a) = \ln(1) = 0 \quad \text{and} \quad f'(a) = \frac11 = 1, $$ and apply the rule.

$$ f(x) \approx f(a) + f'(a) (x-a) \quad \text{for} ~ x \approx a $$ means $$ \ln(x) \approx 0 + 1 (x - 1) = x - 1 \quad \text{for} ~ x \approx 1 $$ or $$ \ln(x + 1) \approx x \quad \text{for} ~ x \approx 0. $$

We could also haven directly chosen $f(x) = \ln(1 + x)$ and $a = 0$, at the price of a slightly harder computation of the derivative, but of course with the same result.

Answer 2

So we know the tangent line will look like $y-y_1=m(x-x_1)$ and we want $y-0=1\cdot(x-0)$.
Since we want our line to go through $(0,0)$, I say find the tangent line at $x=0$.

Answer 3

You are looking at the first degree Maclaurin series of $\ln{(1+x)}$