Limit of $\frac{\tanh(x)-x}{x-\sinh(x)}$

Question

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.

Answer 1

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.

Answer 2

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:

$$\lim_{x\rightarrow 0}\frac{\tanh(x)-x}{x-\sinh(x)}=2$$

Answer 3

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.

Answer 4

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 $2$.

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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}