Find this limit without using L'Hopital's rule or Taylor Series I want to solve $$\lim_{x\to 2}{{\log_2(\sqrt{4x-x^2})-\sqrt{\cos{\pi x}}}\over(\frac{x}{2})^{(\sin{\pi x})}-1}$$
without using differentiation in any context. Basically without using L'Hopital's rule and Taylor's theorem.
I have solved the given limit by using Taylor's theorem in order to find equivalent functions and I have gotten the correct answer. However I am required to do so, without actually differentiating. I am guessing that already established equivalent function formulae and substitutions are the only two tools that I am allowed to use.
Any hints or help will be very much appreciated.
 A: As suggested in comments let's put $x/2=1+t$ so that $$4x-x^2=8(1+t)-4(1+t)^2=4(1-t^2)$$ and thus the expression under limit becomes $$\frac{2+\log_2(1-t^2)-2\sqrt{\cos 2\pi t}}{2((1+t)^{\sin 2\pi t}-1)}$$ Since both terms of denominator tend to $1$ we can safely replace terms by their logarithm (justify this!!) to get $$\frac{2+\log_2(1-t^2)-2\sqrt{\cos 2\pi t}} {2\sin 2\pi t\log(1+t)}\tag{1}$$ This can be split into two terms the first of which is $$\frac{1-\sqrt{\cos 2\pi t}} {\sin 2\pi t\log(1+t)}=\frac{1-\sqrt {\cos 2\pi t}} {1-\cos 2\pi t} \cdot \frac{1-\cos 2\pi t} {(2\pi t) ^2}\cdot\frac{2\pi t} {\sin 2\pi t} \cdot\frac{2\pi t} {\log(1+t)}$$ and this tends to $$A=(1/2)(1/2)(1)(2\pi)=\frac{\pi}{2}$$ The other term in expression $(1)$ is $$\frac{\log_2(1-t^2)}{2\sin 2\pi t\log(1+t)}=\left(-\frac{1}{4\pi\log 2}\right)\cdot\frac{\log(1-t^2)}{(-t^2)}\cdot\frac{2\pi t} {\sin 2\pi t} \cdot\frac{t}{\log(1+t)}$$ which tends to $$B=-\frac{1}{4\pi\log 2}$$ The desired limit is thus $$A+B=\frac{\pi} {2}-\frac{1}{4\pi\log 2}$$
