<s>[INST] Limit $\lim\limits_{x \to π/2}\frac{2x\sin(x) - π}{\cos x}$ without l'hospital There is a problem that is easy to solve with L'hospital but we are required to solve it without it, but I could not find the answer. x* sinx part is especially confusing me, because other examples I have solved did not include such part.
$$\lim_{x \to π/2}\frac{2x\sin(x) - π}{\cos x}$$
 [/INST] Substitute $y=\frac\pi 2 - x$. Then
$$\lim_{x \to π/2}\frac{2x\sin x - π}{\cos x} = \lim_{y\to0}\frac{2(\frac\pi 2-y)\cos y - π}{\sin y}$$
and the function under the limit is
$$\frac{2(\frac\pi 2-y)\cos y - π}{\sin y}
= \frac{(\pi-2y)\cos y - π}{\sin y} \\
= \frac{\pi(\cos y - 1)-2y\cos y}{\sin y} \\
= \pi\frac{\cos y-1}{\sin y}-\frac{2y}{\sin y}\cos y \\
= \pi\frac{\color{red}{\cos y}-\color{blue}1}{\sin y}-2\frac y{\sin y}\cos y \\
= \pi\frac{\color{red}{\cos(2\cdot\frac y2)}-\color{blue}1}{\sin(2\cdot\frac y2)}-2\frac y{\sin y}\cos y \\
= \pi\frac{\color{red}{(\cos^2\frac y2-\sin^2\frac y2)}-\color{blue}{(\sin^2\frac y2+\cos^2\frac y2)}}{2\sin\frac y2\cdot\cos\frac y2}-2\frac y{\sin y}\cos y \\
= \pi\frac{-2\sin^2\frac y2}{2\sin\frac y2\cdot\cos\frac y2}-2\frac y{\sin y}\cos y \\
= -\pi\frac{\sin\frac y2}{\cos\frac y2}-2\frac y{\sin y}\cos y \\
= -\pi\tan\frac y2-2\frac y{\sin y}\cos y
$$
We know (...?) that
$$\tan 0 = 0$$
$$\cos 0 = 1$$
and
$$\lim_{y\to 0}\frac y{\sin y} = 1$$
so the limit sought is
$$\lim_{y\to 0} \left(-\pi\tan\frac y2-2\frac y{\sin y}\cos y\right) = -\pi \cdot 0 - 2\cdot 1 \cdot 1 = \boxed{-2}$$
</s>