Given: $y=\log(1+x)$
Show that $y≈x$ if $x$ gets small (less than 1).
I don't think we're supposed to use Taylor series (because they were never formally introduced in class), but I do think we have to differentiate and show that the derivative of $\log(1+x)$ is approximately equal to $\log(1+x)$ on the interval $0$ to $1$. How should I show this?
By definition of the (natural) logarithm,
$$\log(1+x)=\int_1^{1+x}{du\over u}$$
If $x\approx0$, then ${1\over u}\approx1$ for $1\le u\le 1+x$, in which case
$$\log(1+x)\approx\int_1^{1+x}du=u\Big|_1^{1+x}=(1+x)-1=x$$
(Remark: I wrote $1\le u\le 1+x$ with $x\gt0$ in mind. A more precise version would be $1-|x|\le u\le1+|x|$.)
Use linear approximation around $x=0$. What it means is that in the neighborhood of $x=0$ you are using a tangent line to approximate the actual function.
The tangent line at $x=0$ is given by $y-\ln (1)=f^{'}(0)(x-0)$.
