$\lim\limits_{x\to 0} \frac{x^2\cos x - 6 \ln(1+x^2) +5x^2}{(e^{\sqrt[4]{1+4x^3+8x^4}}-e)\arcsin(x)}$ 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 $\frac{5}{2}$.
 A: With the hint of eyeballfrog, I was able to arrive at the following solution.
We write the nuemreator as $x^2(\cos x - 1) - 6(\ln(1+x^2)-x^2)$ then dividing top and bottom by $x^4$ the limit becomes $$\lim\limits_{x\to 0} \frac{\frac{\cos x - 1}{x^2}-\frac{6}{x^4}(\ln(1+x^2)-x^2)}{\frac{\arcsin x}{x} \cdot \frac{e^{\sqrt{1+4x^3+8x^4}-1}-1}{x^3}} $$
The numerator can be computed using the limits $\lim\limits_{x\to 0}\frac{\cos x -1}{x^2} = -\frac{1}{2}$ and $\lim\limits_{x\to 0} \frac{\ln(1+x^2)-x^2}{x^4} = \frac{-1}{2}$ where the latter is proved in Are all limits solvable without L'Hôpital Rule or Series Expansion
For the bottom, we use the limits in the original post except now it is more divided by $x^3$
$$\lim\limits_{x\to 0} \frac{e^{\sqrt[4]{1+4x^3+8x^4}}-e}{x^3} = e\lim\limits_{x\to 0} \frac{e^{\sqrt[4]{1+4x^3+8x^4}-1}-1}{\sqrt[4]{1+4x^3+8x^4}-1} \cdot \frac{\sqrt[4]{1+4x^3+8x^4}-1}{(1+4x^3+8x^4)-1} \cdot \frac{4x^3+8x^4}{x^3} = e \cdot \frac{1}{4} \cdot 4$$
Combining all of this we arrive at the desired result $\frac{5}{2e}$
A: 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:
$$L=\frac{5}{2e}$$
