$\lim\limits_{x\to 0} \frac{x^2\cos x - 6 \ln(1+x^2) +5x^2}{(e^{\sqrt[4]{1+4x^3+8x^4}}-e)\arcsin(x)}$ without L'Hosptial rule. I have a belief that it is possible to evaluate any limit without using L'Hospital rule or series expansions. However the limit $$\lim\limits_{x\to 0} \frac{x^2\cos x - 6 \ln(1+x^2) +5x^2}{(e^{\sqrt[4]{1+4x^3+8x^4}}-e)\arcsin(x)}$$ stumps me.
I thought at the beginning that is it nothing but dividing top and bottom by $x^2$ and then using the limits: $$\lim\limits_{x\to 0} \frac{\ln(1+x^2)}{x^2} = 1, \quad \quad \lim \limits_{x\to 0} \frac{\arcsin x}{x} = 1$$
And then I found $$\lim\limits_{x\to 0} \frac{e^{\sqrt[4]{1+4x^3+8x^4}}-e}{x} = e\lim\limits_{x\to 0} \frac{e^{\sqrt[4]{1+4x^3+8x^4}-1}-1}{\sqrt[4]{1+4x^3+8x^4}-1} \cdot \frac{\sqrt[4]{1+4x^3+8x^4}-1}{(1+4x^3+8x^4)-1} \cdot \frac{4x^3+8x^4}{x} = 0$$
Which made the limit $\frac{0}{0}$. I have no clue how to split the limit into a product or a sum of existent limits. Is it possible to solve this without L'Hosptial? Is there is possible usage of the squeeze theorem or a geometric interpretation to compute this limit? The answer according to WolframAlpha is $\frac{5}{2e}$ and it is clear how to find the $\frac{1}{e}$ but not the $\frac{5}{2}$.

The numerator is $\sim x^2(\cos(x)-1)+3x^4\sim \frac{5}{2}x^4$
The denominator is $\sim (e^{\sqrt[4]{1+4x^3+8x^4}}-e)x$
So you need to evaluate the limit for
$$\frac{e^{\sqrt[4]{1+4x^3+8x^4}}-e}{x^3}$$
You did $x^2$ instead of $x^3$, so you got $0$, and you can show this term
$$e^{\sqrt[4]{1+4x^3+8x^4}}-e\sim e\cdot x^3$$
Series expansion $\sqrt[4]{1+y}=1+\frac{1}4y+O(y^2)$, substitute $y=4x^3+8x^4$
$$e^{\sqrt[4]{1+4x^3+8x^4}}-e\sim e^{1+x^3+2x^4}-e=e\cdot(e^{x^3+2x^4}-1)\sim e\cdot x^3$$
Final answer:
$$L=\frac{5}{2e}$$
