Curious limits with tanh and sin These two limits can be easily solved by using De l'Hopital Rule multiple times (I think), but I suspect that there could be an easier way... Is there?
\begin{gather}
  \lim_{x\to 0} \frac{\tanh^2 x - \sin^2 x}{x^4} \\
  \lim_{x\to 0} \frac{\sinh^2 x - \tan^2 x}{x^4}
\end{gather}
Thanks for your attention!

Taylor expansions
$$\tanh^2x = x^2-\frac{2x^4}{3}+o\left(x^6\right)$$
$$\sin^2x = x^2-\frac{1}{3}x^4+ o (x^6)$$
$$\lim _{x\to \:0}\:\frac{\tanh ^2\:x\:-\:\sin ^2\:x}{x^4} = \lim _{x\to \:0}\:\frac{x^2-\frac{2x^4}{3}\:-\:\left(x^2-\frac{1}{3}x^4\right)+o(x^6)}{x^4} = \color{red}{-1/3}$$
