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!

From the standard Taylor series expansions, as $x \to 0$,
$$
\begin{align}
\sin x&=x-\frac{x^3}{6}+\frac{x^5}{120}+O(x^6)
\\\tanh x&=x-\frac{x^3}{3}+\frac{2 x^5}{15}+O(x^6)
\end{align}
$$ ones gets
$$
\begin{align}
\left(\sin x\right)^2&=x^2-\frac{x^4}{3}+O(x^6)
\\\left(\tanh x\right)^2&=x^2-\frac{2 x^4}{3}+O(x^6)
\end{align}
$$ giving, as $x \to 0$,

$$
\frac{\left(\tanh x\right)^2-\left(\sin x\right)^2}{x^4}=\frac{-\frac{x^4}{3}+O(x^6)}{x^4}=-\frac13+O(x^2).
$$

