$$ \lim_{x \to 0} \frac{(\sin x - \tanh x)^2}{(e^{x}-1-\ln{(1+x)})^3} \overset{??}{=} \frac{1}{36} $$
If you consult the graph it is intuitive to think that when $ x \to 0 $ the function tends to 1/36. I know that you can apply L'Hospital's rule to solve it, but it's too long. The only solution I see is to apply infinitesimal equivalents, which I have not tried yet.
Thank you all!
On
$$L=\lim_{x \to 0} \frac{(\sin x - \tanh x)^2}{(e^{x}-1-\ln{(1+x)})^3} \overset{??}{=} \frac{1}{36}$$ Expand the den. forsmall values of $x$ it is $$[1+x+x^2/2-1 -x+x^2/2+O(x^3)]=[x^2+O(x^3)]^3,$$ when $|x|<<1$. So the Num. has to be expanded complete upto $x^3$ term using $\sin z= z-z^3/3!+.., \tanh z=z-z^3/3+..$, then $$L=\lim_{x\to 0} \frac{x^6/(36)}{x^6}=\frac{1}{36}.$$
You can use the Taylor expansion to evaluate. In fact, since $$ \sin x=x-\frac16x^3+O(x^5),\tanh x=x-\frac13x^3+O(x^5),e^x-1=x+\frac12x^2+\frac16x^3+O(x^4), \ln(1+x)=x-\frac12x^2+\frac13x^3+O(x^4) $$ then you have $$ \lim_{x \to 0} \frac{(\sin x - \tanh x)^2}{(e^{x}-1-\ln{(1+x)})^3}=\lim_{x \to 0} \frac{\bigg[(x-\frac16x^3)-(x-\frac13x^3)+O(x^5)\bigg]^2}{\bigg[(x+\frac12x^2+\frac16x^3)-(x-\frac12x^2+\frac13x^3)+O(x^4)\bigg]^3}=\frac{\frac{1}{6^2}}{ 1}= \frac{1}{36}. $$