Finding $\lim_{x\to\pi/2}\left(\frac{1-\sin x}{(\pi-2x)^4}\right)(\cos x)(8x^3 - \pi^3)$ using algebra of limits Let $$
F(x) = \left(\frac{1-\sin x}{(\pi-2x)^4}\right)(\cos x)(8x^3 - \pi^3)
$$ 
Then find the limit of $F(x)$ as $x$ tends to $\pi/2$.
How can we find the limit using algebra of limits?
The limit of $\dfrac{1-\sin x}{(\pi-2x)^4}$ as $x$ tends to $\pi/2$ is some non-zero finite number. And the limit of rest part is zero. Then why the limit of the whole function is not $0$?
The limit of $F(x)$ is $-\frac{3\pi^2}{16}$.
How to apply algebra of limits here?

Make life simpler using $x=y+\frac \pi 2$ which makes
$$\lim_{x\to\frac \pi2}\left(\frac{1-\sin (x)}{(\pi-2x)^4}\right)\cos (x)(8x^3 - \pi^3)=\lim_{y\to 0}\frac{\left(4 y^2+6 \pi  y+3 \pi ^2\right) \sin (y) (\cos (y)-1)}{8 y^3}$$ Now, use the usual Taylor the series and you will get
$$\frac{\left(4 y^2+6 \pi  y+3 \pi ^2\right) \sin (y) (\cos (y)-1)}{8 y^3}=-\frac{3 \pi ^2}{16}-\frac{3 \pi  y}{8}+O\left(y^2\right)$$ which shows the limit and how it is approached.
