I've never been 100% happy with using taylors theorem, mainly because I see it used in a bunch of different ways and I'm never sure in which situations it is valid. The way I was introduced to it was that if I have a function $f(x)$ and I want to know its value at points $x$ around $a$ then an approximation is given by $f(x)=f(a)+f'(a)(x-a)+\dots$ where the approximation becomes more accurate the more terms you keep. The more the variable $x$ deviates from the value $a$, the less accurate the expression will be. I'm fine with that. You can then replace the coordinates $x$ with $x+a$, so that $f(x+a)=f(a)+xf'(a)+\dots$, and this is where I start to get confused.
When I see the expression $f(x+a)$ what I think is "given I know a function $f$ at all $x$, then this will tell me the value of $f$ at $x+a$." But this would be useless, as I mean, if we know $f(x)$ at all places, then why would we need an approximation at all? So I'm forced to assume that in the second expression $x$ is fixed and $a$ is variable. So whereas in the first version $x$ represented a variable, it now represents a fixed point.
The reason I'm posting the question is that I'm doing a general relativity course where, in one of the assignments, I'm asked to show that $x^{i}(\lambda, \mu + \delta \mu)-x^i(\lambda, \mu)=\delta \mu \dfrac{\partial{x^i}}{\partial{\mu}} + \dots$ (terms of higher order in $\delta \mu$). In this case, $x^i$ is a 4-position vector, but I guess this could just be any function of two parameters. Obviously I should have learned this stuff prior to doing a G.R. course but I've always been a bit behind. So I'm guessing what's happening here is that the author is only considering $x^i$ as a function of $\mu$ and then taylor expanding the first term? So $x^i(\lambda, \mu + \delta \mu) \cong x^i(\lambda, \mu ) + \left. \dfrac{\partial{x^i(\lambda , \mu + \delta \mu )}}{\partial{\mu}} \right|_{\delta \mu = 0} \delta \mu$, and the first term of this would cancel to give that $x^{i}(\lambda, \mu + \delta \mu)-x^i(\lambda, \mu)= \dfrac{\partial{x^i(\lambda , \mu + \delta \mu )}}{\partial{\mu}}\delta \mu$, evaluated at $\delta \mu =0$. I'm not sure how this turns into the correct answer though, because obviously you can't evaluate it at $0$ and then differentiate, otherwise for example if $f(x)=x+2$ then we would get $f'(0)=0$ when infact $f'(0)=1$.
Apologies for the unclear question, and thanks for any help!
You're thinking about the original Taylor polynomial backwards, I think. When you write $$f(a+x)=f(a)+f'(a)x+\dots,$$ you're thinking of $a$ as fixed and $x$ as variable (and, typically, small). Here we know that $\lim\limits_{x\to 0}\dfrac{\dots}{x} = 0$. Similarly, $$x^i(\lambda,\mu+\Delta\mu) = x^i(\lambda,\mu) + \frac{\partial x^i}{\partial\mu}(\lambda,\mu)\Delta\mu + \dots,$$ where we know that $\lim\limits_{\Delta\mu\to 0}\dfrac{\dots}{\Delta\mu} = 0$. I emphasize that the partial derivative is evaluated at the "base point" $(\lambda,\mu)$, which is, if you like, where we set $\Delta\mu = 0$.