<s>[INST] Compute $\lim_{x\to0} \frac{\cosh x\cosh 2x\cosh 3x \cdots \cosh nx-1}{x^2}$ Compute:
$$\lim_{x\to0} \frac{\cosh x\cosh 2x\cosh 3x \cdots \cosh nx-1}{x^2}$$
How would you tackle this problem? Thanks.
 [/INST] I should want to leave my answer in comment box but my partial answer is too long. I try the following , I know there are a lot of calculations but I think what I've done can help us to find out the answer. 
$\displaystyle \frac{\cosh x\cosh 2x\cosh 3x \cdots \cosh nx-1}{x^2}=\frac{(e^x+e^{-x})(e^{2x}+e^{-2x})\cdot \ldots \cdot(e^{nx}+e^{-nx})-2^n}{2^n \cdot x^2}=$
$$\large \frac{\left(\frac{e^{2x}+1}{e^x}\right)\cdot \left(\frac{e^{4x}+1}{e^{2x}}\right)\cdot \ldots\cdot \left(\frac{e^{2nx}+1}{e^{nx}}\right)-2^{n}}{2^n \cdot x^2}.$$ I want to write this expression beautifully. $e^x =a$. So we have: 
$$\large\frac{\frac{(a^2+1)(a^4+1)\ldots(a^{2n}+1)}{a^{\frac{n(n+1)}{2}}}-2^n}{2^n \cdot x^2}=\frac{(a^2+1)(a^4+1)\ldots(a^{2n}+1)-2^n \cdot a^{\frac{n(n+1)}{2}}}{2^{n}\cdot x^{2} \cdot a^{\frac{n(n+1)}{2}}}. $$
$$=\large \frac{(a^2-1)(a^2+1)(a^4+1)\ldots(a^{2n}+1)-2^n \cdot a^{\frac{n(n+1)}{2}}\cdot(a^2-1)}{2^{n}\cdot x^{2} \cdot a^{\frac{n(n+1)}{2}}\cdot (a^2-1)}$$
$$\large=\frac{(a^{4n}-1)-2^{n}\cdot a^{\frac{n(n+1)}{2}}\cdot(a^2-1)}{2^{n}\cdot x^{2} \cdot a^{\frac{n(n+1)}{2}}\cdot (a^2-1)}$$
$$\large=\frac{e^{4nx}-1-2^{n} \cdot e^{\frac{n(n+1)}{2}\cdot x +2x}+2^{n}\cdot e^{\frac{n(n+1)}{2}\cdot x}}{2^{n} \cdot e^{\frac{n(n+1)}{2}\cdot x +2x} \cdot x^{2} -2^{n}\cdot e^{\frac{n(n+1)}{2}\cdot x} \cdot x^{2}}.$$ 
$$\large \lim_{x \rightarrow 0}{\frac{e^{4nx}-1-2^{n} \cdot e^{\frac{n(n+1)}{2}\cdot x +2x}+2^{n}\cdot e^{\frac{n(n+1)}{2}\cdot x}}{2^{n} \cdot e^{\frac{n(n+1)}{2}\cdot x +2x} \cdot x^{2} -2^{n}\cdot e^{\frac{n(n+1)}{2}\cdot x} \cdot x^{2}}}=$$ and I think this limit can be calculate with L'Hospital. 
Is it ok? 
</s>