Finding $\lim\limits_{x \to 0}\frac{(e^{2\tan(x)}-1) \cdot \ln(2-cos^2(x))}{\sqrt(1+x^3)-(cosx)}$ I have to find the $$\lim_{x \to 0}\frac{(e^{2\tan(x)}-1) \cdot \ln(2-\cos^2(x))}{\sqrt{1+x^3}-(\cos x)}$$ using only notable limits
I managed to solve it having as result $4$ using notable limits, it is quite simple resolve but the solution is not $4$...
I can't find any other way of solution, can somebody solve it using notable limits clearly? (I'm asking many questions regarding the calculus of limit because I'm preparing for a limit's test)

$$\lim_{x \to 0}\frac{(e^{2\tan x}-1) \cdot \ln(2-\cos^2x)}{\sqrt{1+x^3}-\cos x}=\lim_{x \to 0}\frac{e^{2\tan x}-1}{2\tan x}\cdot\lim_{x \to 0}\frac{ \ln(1+\sin^2x)}{\sin^2x}\cdot\lim_{x \to 0}\frac{2\tan x\cdot\sin^2x}{\sqrt{1+x^3}-\cos x}=2\cdot1\cdot\lim_{x \to 0}\frac{\tan x}{x}\lim_{x \to 0}\cdot\frac{\sin^2 x}{x^2}\cdot\lim_{x \to 0}\frac{x^3}{\sqrt{1+x^3}-1+(1-\cos x)}=2\lim_{x \to 0}\frac{x^3}{\frac{x^3}{\sqrt{1+x^3}+1}+(1-\cos x)}= 2\lim_{x \to 0}\frac{1}{\frac{1}{\sqrt{1+x^3}+1}+\frac{1-\cos x}{x^3}}=\frac{2}{\frac{1}{2}+\lim_{x \to 0}\frac{1-\cos x}{x^2}\cdot\frac{1}{x}}=\frac{2}{\frac{1}{2}+\frac{1}{2}\cdot\lim_{x \to 0}\frac{1}{x}}.$$ 
 If we see it as the lateral limit, then the limit is equal to $0.$
