I cannot work out how i would start this question.
Use the Maclaurin series to evaluate the limit
lim x →0 ((x cos x − x)/sin^3x)
Any help is much appreciated
Thank you once again
ViperRobK
$\displaystyle \lim_{x\to 0}\frac{xcos(x)-x}{sin^{3}(x)}$
Use the respective known series for sine and cosine.
$\displaystyle sin(x)=\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k+1}}{(2k+1)!}$
$\displaystyle xcos(x)=\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k+1}}{(2k)!}$
The first few terms:
$\displaystyle \lim_{x\to 0}\left[\frac{\frac{-x^{3}}{2!}+\frac{x^{5}}{4!}-\frac{x^{7}}{6!}+\frac{x^{9}}{8!}-.................}{x^{3}-\frac{x^{5}}{2}+\frac{13x^{7}}{120}-\frac{41x^{9}}{3024}+............}\right]$