Finding $\lim_{x \to 0} \left( \frac{\sin(x)-3\sinh(x)+2x}{x^2(\tanh(2x)+ \sin(x))} \right)$ I am trying to evaluate the following limit:
$$L=\lim_{x \to 0} \left( \frac{\sin(x)-3\sinh(x)+2x}{x^2(\tanh(2x)+\sin(x))} \right)$$
Begin by rewriting the limit as:
$$L=\frac{\lim\limits_{x \to 0}\left(\cfrac{\sin(x)-3\sinh(x)+2x}{x^2} \right)}{\lim\limits_{x \to 0}(\tanh(2x)+\sin(x))} \tag{1}$$
Applying L'Hospital's Rule to the numerator only:
$$L=\frac{\lim\limits_{x \to 0}\left(\cfrac{\cos(x)-3\cosh(x)+2}{2x} \right)}{\lim\limits_{x \to 0}(\tanh(2x)+\sin(x))} \tag{2}$$
The numerator is still in an indeterminate form, applying L'Hopital to the numerator again:
$$L=\frac{\lim\limits_{x \to 0}\left(\cfrac{-\sin(x)-3\sinh(x)}{2} \right)}{\lim\limits_{x \to 0}(\tanh(2x)+\sin(x))} \tag{3}$$
Rewriting as a single limit:
$$L=-\frac{1}{2}\lim_{x \to 0}\frac{\sin(x)+3\sinh(x)}{\tanh(2x)+\sin(x)} \tag{4}$$
And applying  L'Hospital's Rule...
$$L=-\frac{1}{2}\lim_{x \to 0}\left(\frac{\cos(x)+3\cosh(x)}{2\operatorname{sech}^2(2x)+\cos(x)} \right)=-\frac{2}{3} \tag{5}$$
But according to Wolfram Alpha, $L=-\frac{2}{9}$
So something must be wrong in my calculation (I guess it's the limit of a product bit)?
 A: We can't separate the limit in this way and then apply l'Hospital's rule only to a single part.
In this case we can proceed as follows
$$\frac{\sin(x)-3\sinh(x)+2x}{x^2(\tanh(2x)+\sin(x))}= \frac{x}{\tanh(2x)+\sin(x)} \frac{\sin(x)-3\sinh(x)+2x}{x^3}$$
and use standard limit for this one
$$ \frac{x}{\tanh(2x)+\sin(x)} =  \frac{1}{2\frac{\tanh(2x)}{2x}+\frac{\sin(x)}{x}} $$
and then apply l'Hospital's rule for the second part to obtain the result according to the product rule
$$\lim_{x\to x_0} f(x)g(x)=\lim_{x\to x_0} f(x)\cdot \lim_{x\to x_0} g(x)$$
A: $$\sinh x=x+\frac{x^3}{6}+O\left(x^4\right)$$
$$\tanh(2x)=2 x-\frac{8 x^3}{3}+O\left(x^4\right)$$
$$\sin x=x-\frac{x^3}{6}+O\left(x^4\right)$$
The limit can be rewritten as
$$\frac{x-\frac{x^3}{6}-3 \left(\frac{x^3}{6}+x\right)+2 x}{x^2 \left(2x-\frac{8 x^3}{3}+x-\frac{x^3}{6}\right)}\to -\frac{2}{9}\text{ as }x\to 0$$
A: To make it as simple as possible, I use equivalents  via Taylor-Young's formula at the relevant order:

*

*$\sin x-2\sinh x+2x=x-\frac{x^3}6+o(x^3)-3x-\frac{3x^3}6+o(x^3)+2x=-\frac{2x^3}3+o(x^3)$.

Therefore $\:\sin x-2\sinh x+2x\sim_0-\dfrac{2x^3}3.$

*

*$\tanh 2x+\sin x=2x+o(x)+x+o(x)=3x+o(x)\sim_0 3x$
As a consequence
$$\frac{\sin x-2\sinh x+2x}{x^2(\tanh 2x+\sin x)}\sim_0\frac{-\cfrac{2x^3}3}{x^2\cdot 3x}=-\frac29.$$
