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!
 A: You could use power series:
$$\eqalign{\sinh x-\sin x\cosh x
  &=\Bigl(x+\frac{x^3}6+\cdots\Bigr)
    -\Bigl(x-\frac{x^3}6+\cdots\Bigr)\Bigl(1+\frac{x^2}2+\cdots\Bigr)\cr
  &=-\frac{x^3}6+\cdots\cr}$$
and so
$$\frac{\tanh x-\sin x}{x^3}
  =\frac1{\cosh x}\frac{\sinh x-\sin x\cosh x}{x^3}\to-\frac16\ ;$$
similarly
$$\sinh x+\sin x\cosh x=2x+\cdots$$
so
$$\frac{\tanh x+\sin x}{x}
  =\frac1{\cosh x}\frac{\sinh x+\sin x\cosh x}{x}\to2\ ;$$
now multiply these.
A: 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 
$$\frac{\sinh^2 x-\tan^2x}{x^4}=\frac{-\dfrac{x^4}3+o(x^4)}{x^4}\to-\frac13.$$

A: 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 $x \to 0$,

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

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