# Solving an equation involving a limit for two variables

I've been attempted to solve the following equation for values of a and c that would make it true:

$$\lim_{x\to0}{\sqrt{ae^{2x+1}+cx\arccos{x}}-3e^x\over{\sin{2x}}}=1$$

So far I've rationalized the numerator, resulting in the following: $$\lim_{x\to0}{ae^{2x+1}+cx\arccos{x}-9e^{2x}\over{\sin{2x}(\sqrt{ae^{2x+1}+cx\arccos{x}}+3e^x})}=1$$

However, no matter what I do from this point onward, I cannot find a way to factor a $2x$ or $\sin{2x}$ out of the numerator, which would allow me to break the limit apart and handle the $\sin{2x}$ in the denominator. Any ideas for what I should do at this stage?

Note: the use of L'Hopital's rule is not permitted within the bounds of the question.

First of all, since $\lim_{x\to0}\sin2x=0$, the numerator must have limit $0$. Thus $$\sqrt{ae}-3=0$$ and therefore $ae=9$. So, after rationalizing, you have $$\lim_{x\to0}\frac{cx\arccos x} {\sin 2x(\sqrt{9e^{2x}+cx\arccos x}+3e^x)}= \frac{1}{2}\frac{c\frac{\pi}{2}}{\sqrt{9}+3}$$ by pairing the $x$ in the numerator with $\sin2x$ in the denominator.
$$1=\lim_{x\to0}{ae^{2x+1}+cx\arccos{x}-9e^{2x}\over{\sin{2x}(\sqrt{ae^{2x+1}+cx\arccos{x}}+3e^x})}=\frac{1}{\sqrt{ae}+3}\lim_{x\to 0}\left[\frac{(ae-9)e^{2x}}{\sin 2x}+\frac{c\arccos{x}}{2}\frac{2x}{\sin2x}\right]$$
Using the standard Taylor series, the term under the radical $$a e^{2 x+1}+c x \cos ^{-1}(x)=e a+x \left(2 e a+\frac{\pi c}{2}\right)+O\left(x^2\right)$$ that is to say $$\sqrt{a e^{2 x+1}+c x \cos ^{-1}(x)}=\sqrt{ea}\sqrt{1+\frac{x \left(2 e a+\frac{\pi c}{2}\right)}{e a}+O\left(x^2\right)}$$ Now, remembering that $$\sqrt{1+t}= 1+\frac{t}{2}+O\left(t^2\right)$$ makes the numerator to be $$\sqrt{a e^{2 x+1}+c x \cos ^{-1}(x)}-3e^{x}=\left(\sqrt{ae} -3\right)+x \left(\frac{2 e a+\frac{\pi c}{2}}{2 \sqrt{ae} }-3\right)+O\left(x^2\right)$$ The denominator being $$\sin(2x)=2 x+O\left(x^2\right)$$ then $$\frac{\sqrt{a e^{2 x+1}+c x \cos ^{-1}(x)}-3e^{x}}{\sin(2x)}=\frac{\left(\sqrt{ae} -3\right)+x \left(\frac{2 e a+\frac{\pi c}{2}}{2 \sqrt{ae} }-3\right)+O\left(x^2\right) }{2 x+O\left(x^2\right) }$$ This first implies $ae=9$ and if you want the limit to be $L$, then $$\frac{1}{2} \left(\frac{2 e a+\frac{\pi c}{2}}{2 \sqrt{ae} }-3\right)=L\implies \frac{\pi c}{24}=L\implies c=\frac{24L}\pi$$