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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)?

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  • $\begingroup$ Please do not make your subject lines consist entirely and exclusively of math formulas; this prevents many of the usual right-click actions on them. And also, use \lim, \cos, \cosh, \sin, \tan, etc. If one of them doesn't work, you can use \mathop{sech} or \mathrm{sech}. The function names and the limits should not be typeset in math italic. $\endgroup$ – Arturo Magidin Nov 2 '20 at 18:09
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    $\begingroup$ (1) is invalid. You cannot rewrite a limit of a quotient as a quotient of limits when the limit of the denominator is $0$. That is not a valid limit law. $\endgroup$ – Arturo Magidin Nov 2 '20 at 18:10
  • $\begingroup$ @Arturo Magidin Thanks. What about when the denominator is of indeterminate forms e.g.: $\lim_{x \to 0}\ \left(\frac{\sin(x)-3\sinh(x)+2x}{\sin(x)x^2\left(\frac{\tanh(2x)}{\sin(x)}+1) \right)} \right)=\frac{ \lim_{x \to 0}\left(\frac{\sin(x)-3\sinh(x)+2x}{sin(x)x^2} \right)}{\lim_{x \to 0}\left(\frac{\tanh(2x)}{\sin(x)+1} \right)}$? $\endgroup$ – Chern Simons Nov 2 '20 at 18:30
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    $\begingroup$ You still can't. That's not a valid limit law. The limit of the quotient is equal to the quotient of the limits if they both exist and the limit of the denominator is not zero. In any other situation there is no guarantee that whatever manipulations you make will lead you to a correct answer. $\endgroup$ – Arturo Magidin Nov 2 '20 at 18:33
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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

$$\lim_{x\to x_0} f(x)g(x)=\lim_{x\to x_0} f(x)\cdot \lim_{x\to x_0} g(x)$$

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

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

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