Linear approximation $y= \ln(1+x)$ for small x 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.
 A: 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.
A: 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
 $x=0$.
A: You are looking at the first degree Maclaurin series of $\ln{(1+x)}$
