Limit of $\frac{\tanh(x)-x}{x-\sinh(x)}$ What is the limit of $$\frac{\tanh(x)-x}{x-\sinh(x)}$$
when $x$ go to $0$?
I calculated so far using L'Hopital's rule and transformed into this.
$$\frac{1-\tanh(x)}{(\tanh(x))(1-\cosh(x))}$$
but I have no idea what would be the next step...
Can someone teach me how to find this limit? Thank you in advance.
 A: For this kind of problems, Taylor series are really useful.
Rememeber that
$$\tanh(x)=x-\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$
$$\sinh(x)=x+\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$ which make
$$\frac{\tanh(x)-x}{x-\sinh(x)}=\frac{-\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right) } {-\frac{x^3}{6}-\frac{x^5}{120}+O\left(x^7\right) }=2-\frac{9 x^2}{10}+O\left(x^4\right)$$ showing the limit and also how it is approached.
A: Sometimes you need to apply L'Hôpital's rule more than once to get a result with a limit, as long as you get the indeterminate $\frac{0}{0}$ at each stage. Also, you made a mistake with your calculations. Note that
$$\frac{d}{dx}\sinh x = \cosh x \tag{1}\label{eq1}$$
$$\frac{d}{dx}\cosh x = \sinh x \tag{2}\label{eq2}$$
$$\frac{d}{dx}\tanh x = 1 - \tanh^2 x \tag{3}\label{eq3}$$
Thus, your numerator would be $1 - \tanh^2 x - 1 = -\tanh^2 x$ and the denominator would be $1 - \cosh x$. The following uses L'Hôpital's rule at each of the first $3$ lines shown below, with the limit being $\frac{0}{0}$ for the first $2$ times:
\begin{align}
\lim_{x \to 0} \frac{\tanh(x)-x}{x-\sinh(x)} &= \lim_{x \to 0} \frac{-\tanh^2 x}{1 - \cosh x} \\
& = \lim_{x \to 0} \frac{-2\tanh x(1 - \tanh^x)}{-\sinh} = \lim_{x \to 0} \frac{-2\tanh x + 2\tanh^3 x}{-\sinh x} \\
& = \lim_{x \to 0} \frac{-2(1 - \tanh^2 x) + 6\tanh^2 x(1 - \tanh^2 x)}{-\cosh x} \\
& = \frac{-2}{-1} = 2 \tag{4}\label{eq4}
\end{align}
The final limit comes from using that $\tanh(0) = 0$ and $\cosh(0) = 1$. As you can see, you need to use L'Hôpital's rule $3$ times for this problem.
A: If you want to stick to l'Hopital's...
Applying it once yields:
$$\frac{\text{sech}^2(x)-1}{1-\cosh(x)}$$
which still gives $\frac{0}{0}$ at $x=0$, so we apply it again:
$$\frac{-2\tanh(x)\text{sech}^2(x)}{-\sinh(x)}$$
which still doesn't work. Cancelling the negative sign and doing it again (this time with product rule in the numerator!):
$$\frac{2\text{sech}^2(x)\Big(\text{sech}^2(x)-2\tan^2(x)\Big)}{\cosh(x)}$$
Using the fact that $\text{sech}^2(0)=1$, $\tanh(0)=0$, and $\cosh(0)=1$, you get that:
$$\lim_{x\rightarrow 0}\frac{\tanh(x)-x}{x-\sinh(x)}=2$$
A: After the first application of l'Hopital where you get
$$
\frac{\frac1{\cosh^2x}-1}{1-\cosh x}
$$
you can continue purely algebraically by extracting factors, applying binomial formulas and cancelling under the assumption $x\ne 0$
$$
=\frac1{\cosh^2x}\frac{(1-\cosh x)(1+\cosh x)}{1-\cosh x}=\frac{1+\cosh x}{\cosh^2x}.
$$
This is continuous at $x=0$ and can be directly evaluated there to the value $2$.

It can also be helful to simplify (using the multiplication law of the limit) before the application of l'Hopital, to avoid the occurrence of "arcane" derivatives
\begin{align}
\lim_{x\to 0}\frac{\tanh x-x}{x-\sinh x}
&=\frac1{\cosh 0}\cdot\lim_{x\to 0}\frac{\sinh x-x\cosh x}{x-\sinh x}
\\
&\overset{\rm l'Hop}=\lim_{x\to 0}\frac{\cosh x-\cosh x-x\sinh x}{1-\cosh x}
\\
&=(1+\cosh 0)⋅\lim_{x\to 0}\frac{x\sinh x}{\cosh^2x-1}=2\lim_{x\to 0}\frac{x}{\sinh x}=2
\end{align}
