$\lim\limits_{t\to 0}(\frac{2+\cos2t}{3\sin^4t}-\frac{1}{t^4})$ It's from this answer.
https://math.stackexchange.com/a/368574/481435
I cannot evaluate the last limit.
I arranged that limit to slightly clearer form.
$$\lim\limits_{t\to 0}\left(\frac{2+\cos2t}{3\sin^4t}-\frac{1}{t^4}\right)=\frac{1}{45}$$
I have no idea how to confirm that value. I know in theory I can use L'hospital's theorem but it would be endless derivatives for this problem. I tried Taylor expansion but I failed.

Note that
$$\frac{2+\cos2t}{3\sin^4t}-\frac{1}{t^4}=\frac{2t^4+t^4\cdot\cos2t-3\sin^4t}{3t^4\cdot\sin^4t}$$
thus we need to expand in such way to have $8^{th}$ order terms, then


*

*$\cos2t=1-\frac{4t^2}{2}+\frac{16t^4}{24}+o(t^4)=1-2t^2+\frac{2t^4}{3}+o(t^4)$

*$\sin t=t-\frac{t^3}{6}+o(t^3)$

*$\sin^4
   t=\left(t-\frac{t^3}{6}+o(t^3)\right)^4=t^4-\frac{2t^6}{3}+\frac{t^8}{5}+o(t^8)$


then
$$\frac{2t^4+t^4\cdot\cos2t-3\sin^4t}{3t^4\cdot\sin^4t}=\frac{2t^4+t^4-2t^6+\frac{2t^8}{3}+o(t^8)-3t^4+2t^6-\frac{3t^8}{5}+o(t^8)}{3t^8+o(t^8)}=\frac{\frac{t^8}{15}+o(t^8)}{3t^8}=\frac1{45}+o(1)\to \frac1{45}$$
