Limit goes to infinity question: $\lim _{x\to \infty }\left(x^6\left[e^{-\frac{1}{2x^3}}-\cos\left(\frac{1}{x\sqrt{x}}\right)\right]\right)$

Question

$\lim _{x\to \infty }\left(x^6\left[e^{-\frac{1}{2x^3}}-\cos\left(\frac{1}{x\sqrt{x}}\right)\right]\right)$
Any tip on how to calculate it?

The solution is :$\frac{1}{12}$ I dont need the way to solution, just a tip on how to... I have tried Taylor series, it wont work for me, I will get something really ugly and at the end, I am still not managing to get rid of the $x^6$. I tried doing taylor series of $e^x$, then instead of $x$ I put $\frac{1}{2x^3}$, but still wont work to me... any tip will be really favoured.

Answer 1

In such cases, I always find that transforming the beast into a limit at zero is easier. Perform the substitution $$ x^{3/2}=1/t $$ so $x^3=1/t^2$ and $x^6=1/t^4$. Now the limit becomes $$ \lim_{t\to0^+}\frac{e^{-t^2/2}-\cos t}{t^4} $$ and the Taylor expansion of the numerator up to degree $4$ is $$ 1-\frac{t^2/2}{1!}+\frac{t^4/4}{2!}-1+\frac{t^2}{2!}-\frac{t^4}{4!}+o(t^4)=\frac{t^4}{12}+o(t^4) $$