<s>[INST] 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)?
 [/INST] $$\sinh x=x+\frac{x^3}{6}+O\left(x^4\right)$$
$$\tanh(2x)=2 x-\frac{8 x^3}{3}+O\left(x^4\right)$$
$$\sin x=x-\frac{x^3}{6}+O\left(x^4\right)$$
The limit can be rewritten as
$$\frac{x-\frac{x^3}{6}-3 \left(\frac{x^3}{6}+x\right)+2 x}{x^2 \left(2x-\frac{8 x^3}{3}+x-\frac{x^3}{6}\right)}\to -\frac{2}{9}\text{ as }x\to 0$$
</s>