<s>[INST] 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!
 [/INST] There is: Taylor's formula at order $4$:


*

*$\tanh x=x-\dfrac{x^3}3+o(x^4)$, hence
$$\tanh^2 x=\Bigl(x-\dfrac{x^3}3+o(x^4)\Bigl)^2=x^2-\dfrac{2x^4}3+o(x^4).$$

*$\sin^2x=\frac12(1-\cos 2x)=\frac12\Bigl(1-1+\dfrac{4x^2}2-\dfrac{16x^4}{24}+o(x^4)\Bigr)=x^2-\dfrac{x^4}3+o(x^4)$.
Thus 
$$\frac{\tanh^2 x-\sin^2x}{x^4}=\frac{-\dfrac{x^4}3+o(x^4)}{x^4}=-\frac13+o(1)\to-\frac13.$$


Similarly:


*

*$\tan x=x+\dfrac{x^3}3+o(x^4)$, hence
$\tan^2 x=x^2+\dfrac{2x^4}3+o(x^4).$

*$\sinh^2x=\frac12(\cosh 2x-1)=\frac12\Bigl(1+\dfrac{4x^2}2+\dfrac{16x^4}{24}+o(x^4)-1\Bigr)=x^2+\dfrac{x^4}3+o(x^4)$.
Thus 
$$\frac{\sinh^2 x-\tan^2x}{x^4}=\frac{-\dfrac{x^4}3+o(x^4)}{x^4}\to-\frac13.$$

</s>