Using L'Hopital's rule to find the limit of this function I need to find the limit of this function using L'Hopitals rule and I am not sure about my calculations
$$ \lim_{x\to (-\infty)}\frac{e^\frac{1}{x^3}-1}{\arctan\frac{2}{x^4}}$$
Using Lhopital rule and finding the derivative I get
$$ \lim_{x\to (-\infty)}\frac{e^\frac{1}{x^3}*\frac{-3}{x^4} }{\frac{1}{1+\frac{4}{x^8}}* \frac{-8}{x^5}}$$
but Im not sure how to proceed further I tried simplifying but Im not 100% confident that's correct
$$ \lim_{x\to (-\infty)}\frac{e^\frac{1}{x^3}*\frac{-3}{x^4} }{\frac{1}{1+\frac{4}{x^8}}* \frac{-8}{x^5}} = \frac{e^\frac{1}{x^3}*\frac{-3}{1} }{\frac{1}{1+\frac{4}{x^8}}* \frac{-8}{x}} = \frac{e^\frac{1}{x^3}*-3 }{\frac{-8}{(1+\frac{4}{x^8})x}} = \frac{e^\frac{1}{x^3}*-3 }{\frac{-8}{x+\frac{4}{x^7}}} = e^\frac{1}{x^3}*-3  *{\frac{x+\frac{4}{x^7}}{-8}} = 1*-3 {\frac{-\infty+0}{-8}} = -\infty$$
Hint:
We don't actually need L'Hopital's rule
Set $-\dfrac1x=h\implies h\to0^+$
$$\lim_{h\to0^+}\dfrac{e^{-h^3}-1}{\arctan(2h^4)}=\lim_{h\to0^+}\dfrac{e^{-h^3}-1}{-h^3}\cdot\dfrac1{\lim_{h\to0^+}\dfrac{\arctan(2h^4)}{2h^4}}\cdot\lim_{h\to0^+}\dfrac{-h^3}{2h^4}=?$$
Hint:
We don't actually need L'Hopital's rule
Set $-\dfrac1x=h\implies h\to0^+$
$$\lim_{h\to0^+}\dfrac{e^{-h^3}-1}{\arctan(2h^4)}=\lim_{h\to0^+}\dfrac{e^{-h^3}-1}{-h^3}\cdot\dfrac1{\lim_{h\to0^+}\dfrac{\arctan(2h^4)}{2h^4}}\cdot\lim_{h\to0^+}\dfrac{-h^3}{2h^4}=?$$