# 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.

• My first instinct is to substitute $y=x/2-1$. Mar 13, 2021 at 22:41
• It is hard to answer this without knowing what the "already established equivalent function formulae" are.
– Ted
Mar 13, 2021 at 23:23

The method is to decompose this function into several whose limits are easier.

First, substituting $$x = 2+y$$ (to better see what's going on) we have $$\lim_{x\to 2} \frac{\log_2(\sqrt{4x-x^2}) - \sqrt{\cos\pi x}}{(\frac{x}{2})^{\sin \pi x} - 1} = \lim_{y\to 0} \frac{\log_2\sqrt{4-y^2}-\sqrt{\cos \pi y}}{(1+\frac{y}{2})^{\sin\pi y} - 1}$$ We have then $$\frac{\log_2\sqrt{4-y^2}-\sqrt{\cos \pi y}}{(1+\frac{y}{2})^{\sin\pi y} - 1} = \frac{\log_2\sqrt{4-y^2}-\sqrt{\cos \pi y}}{y^2} \frac{y^2}{(1+\frac{y}{2})^{\sin\pi y} - 1}$$ $$\frac{\log_2\sqrt{4-y^2}-\sqrt{\cos \pi y}}{y^2} = \frac{\log_2\sqrt{4-y^2}-1}{y^2} + \frac{1-\sqrt{\cos \pi y}}{y^2}$$ \begin{align} \frac{\log_2\sqrt{4-y^2}-1}{y^2} &=\frac{\log_2\sqrt{4-y^2}-\log_2 2}{y^2} =\frac{\log_2(\frac12\sqrt{4-y^2})}{y^2}= \frac{\log_2\sqrt{1-\frac{y^2}{4}}}{y^2} = \\ &= \frac{\log_2\sqrt{1-\frac{y^2}{4}}}{\sqrt{1-\frac{y^2}{4}} - 1} \cdot \frac{\sqrt{1-\frac{y^2}{4}} -1}{y^2} =\\ &= \frac{\log_2\sqrt{1-\frac{y^2}{4}}}{\sqrt{1-\frac{y^2}{4}} - 1} \cdot \frac{(1-\frac{y^2}{4})-1}{y^2(\sqrt{1-\frac{y^2}{4}} +1)} = \\ &= \frac{1}{\ln 2}\frac{\ln\sqrt{1-\frac{y^2}{4}}}{\sqrt{1-\frac{y^2}{4}} - 1} \cdot \frac{-\frac14}{\sqrt{1-\frac{y^2}{4}} +1} \end{align} We have $$\lim_{y\to 0} \frac{-\frac14}{\sqrt{1-\frac{y^2}{4}} +1} = \frac{-\frac14}{1+1} = -\frac18$$ and using $$t = \sqrt{1-\frac{y^2}{4}} - 1$$ we have $$\lim_{y\to 0} \frac{\ln\sqrt{1-\frac{y^2}{4}}}{\sqrt{1-\frac{y^2}{4}} - 1} = \lim_{t\to 0} \frac{\ln(1+t)}{t} = 1$$ (assuming we already know this limit), so $$\lim_{y\to 0} \frac{\log_2\sqrt{4-y^2}-1}{y^2} = -\frac{1}{8\ln 2}$$ Similarily, we have \begin{align} \frac{1-\sqrt{\cos \pi y}}{y^2} &= \frac{1-\cos \pi y}{y^2(1+\sqrt{\cos \pi y})} = \frac{2\sin^2 \frac{\pi y}{2}}{y^2(1+\sqrt{\cos \pi y})} = \\ &= \left(\frac{\sin\frac{\pi y}{2}}{\frac{\pi y}{2}}\right)^2 \frac{\pi^2}{2(1+\sqrt{\cos \pi y})}\end{align} Assuming we know that $$\lim_{t\to 0}\frac{\sin t}{t} =1$$, we get $$\lim_{y\to 0} \frac{1-\sqrt{\cos \pi y}}{y^2} = 1^2 \frac{\pi^2}{2(1+\sqrt{1})} = \frac{\pi^2}{4}$$ Finally we have \begin{align} \frac{y^2}{(1+\frac{y}{2})^{\sin\pi y} - 1} &= \frac{y^2}{e^{(\ln(1+\frac{y}{2}))(\sin\pi y)} - 1} = \frac{y^2}{(\ln(1+\frac{y}{2}))(\sin\pi y)}\frac{(\ln(1+\frac{y}{2}))(\sin\pi y)}{e^{(\ln(1+\frac{y}{2}))(\sin\pi y)} - 1} = \\ &= \frac{2}{\pi}\frac{y/2}{\ln(1+\frac{y}{2})}\frac{\pi y}{\sin\pi y}\frac{(\ln(1+\frac{y}{2}))(\sin\pi y)}{e^{(\ln(1+\frac{y}{2}))(\sin\pi y)} - 1}\end{align} Assuming we know that $$\lim_{t\to 0}\frac{t}{\ln(1+t)} = \lim_{t\to 0} \frac{t}{\sin t} = \lim_{t\to 0}\frac{t}{e^t-1} = 1$$ we get $$\lim_{y\to 0} \frac{y^2}{(1+\frac{y}{2})^{\sin\pi y} - 1} = \frac{2}{\pi}\cdot 1\cdot 1\cdot 1 = \frac{2}{\pi}$$ In total, we get $$\lim_{y\to 0} \frac{\log_2\sqrt{4-y^2}-\sqrt{\cos \pi y}}{(1+\frac{y}{2})^{\sin\pi y} - 1} = (-\frac{1}{8\ln 2}+\frac{\pi^2}{4}) \frac{2}{\pi}$$

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 $$A+B=\frac{\pi} {2}-\frac{1}{4\pi\log 2}$$