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!

First get rid of the squares with
$$\lim_{x\to 0} \frac{\tanh^2 x - \sin^2 x}{x^4}=\lim_{x\to 0} \frac{(\tanh x - \sin x)(\tanh^2 x + \sin^2 x)}{x^4}=2\lim_{x\to 0} \frac{\tanh x - \sin x}{x^3}.$$
Then, as the functions are odd, there will be no quadratic term, and you can substitute $x=\sqrt t$ to skip it. Then by two applications of L'Hospital
$$2\lim_{t\to 0} \frac{\tanh\sqrt t - \sin\sqrt t}{t\sqrt t}=2\lim_{t\to 0}\frac{(\tanh^2\sqrt t-1)-\cos\sqrt t}{2\sqrt t\frac32\sqrt t}\\
=2\lim_{t\to 0}\frac{2\tanh\sqrt t(\tanh^2\sqrt t-1)+\sin\sqrt t}{2\sqrt t\,3}=-\frac13.$$
By the substitution $x\leftrightarrow ix$, the two limits are equal.
