How would I calculate $\lim_{x \rightarrow 0}{\frac{\int_{0}^{x} \frac{\sin(t)}{t}-\tan(x)}{2x(1-\cos(x))}}$ using a Maclaurin polynomial? For the integral in the numerator, the polynomial doesn't seem to exist beyond degree 1.
Thanks!
How would I calculate $\lim_{x \rightarrow 0}{\frac{\int_{0}^{x} \frac{\sin(t)}{t}-\tan(x)}{2x(1-\cos(x))}}$ using a Maclaurin polynomial? For the integral in the numerator, the polynomial doesn't seem to exist beyond degree 1.
Thanks!
There is no trouble with the integral. Just write down the expansion of $\sin t$, divide by $t$, and integrate term by term.
In detail, $\sin t$ has expansion $t-t^3/6+\cdots$. Divide by $t$. We get $1-t^2/6+\cdots$. Integrate from $0$ to $x$. We get $x-x^3/18+\cdots$. If you are worried about what happens at $0$, don't. The limit is $1$. Anyway, the integral is not sensitive to the value of the function at just one point.
You will need a bit of the expansion of $\tan x$. Possibly you could just compute the terms (you will need up to the $x^3$ term) from the definition. You will need the first three derivatives of $\tan x$. There are other ways, like working with $\sin x$ and $\cos x$.
We have, for $x$ near zero, $ \sin t = t-t^3/6+o(t^5), \ \ \tan x =x+x^3/3+o(x^5),\ \ \cos x =1-x^2/2+o(x^4). $ Then $ \frac{\int_0^x\frac{\sin t}t dt-\tan x}{2x(1-\cos x)} =\frac{\int_0^x(1-t^2/6+o(t^4))dt-(x+x^3/3+o(x^5))}{2x(x^2/2+o(x^4))} =\frac{x-x^3/18+o(x^5)-(x+x^3/3+o(x^5))}{2x(x^2/2+o(x^4))} =\frac{-7x^3/18+o(x^5)}{x^3+o(x^5))} =\frac{-7/18+o(x^2)}{1+o(x^2))} \to -\frac7{18}. $