Curious limits with tanh and sin

Question

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!

Answer 1

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.

Answer 2

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 $$\frac{\sinh^2 x-\tan^2x}{x^4}=\frac{-\dfrac{x^4}3+o(x^4)}{x^4}\to-\frac13.$$

Answer 3

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 $x \to 0$,

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

Answer 4

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

Answer 5

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.