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?
 A: The limit of $\frac{1-\sin x}{(\pi-2x)^4}$ as $x$ tends to $\pi/2$ is NOT some non zero finite number.
Note that $(8x^3 - \pi^3)=(2x-\pi)(4x^2+2x\pi+\pi^2)$, then 
$$F(x)=\frac{1-\sin x}{(2x-\pi)^2}\cdot\frac{\cos x}{(2x-\pi)}\cdot (4x^2+2x\pi+\pi^2)$$
Now evaluate the limit of each factor. This time they are all finite numbers and we can apply the rule related to your previous question How to apply algebra of limits?
Can you take it from here?
A: 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.
A: Because we are dealing with an indeterminate form $0\cdot \infty$ and algebraic theorems do not apply.
To solve the limit let $y=\pi/2-x\to 0$ then
$$\frac{1-\sin x}{(\pi-2x)^4}\cos x(8x^3 - \pi^3)=-\frac{1-\cos y}{16y^4}8y\sin y \left((\pi/2-y)^2+(\pi/2-y)\pi/2+(\pi/2)^2\right)=$$
$$=-\frac{1-\cos y}{2y^2}\frac{\sin y}y \left((\pi/2-y)^2+(\pi/2-y)\pi/2+(\pi/2)^2\right)$$
