Calculate $\lim_{x \rightarrow 0}{\frac{\int_{0}^{x} \frac{\sin(t)}{t}-\tan(x)}{2x(1-\cos(x))}}$ using Mclaurin polynomial 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!

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 
$$
\frac{\int_0^x\frac{\sin t}t dt-\tan x}{2x(1-\cos x)}
=\frac{\int_0^x(1-t^2/6+o(t^4))dt-(x+x^3/3+o(x^5))}{2x(x^2/2+o(x^4))}
=\frac{x-x^3/18+o(x^5)-(x+x^3/3+o(x^5))}{2x(x^2/2+o(x^4))}
=\frac{-7x^3/18+o(x^5)}{x^3+o(x^5))}
=\frac{-7/18+o(x^2)}{1+o(x^2))}
\to -\frac7{18}.
$$
