Mathematics Asked by Toni Rivera Vargas on December 25, 2020

$$ lim_{x to 0} frac{(sin x – tanh x)^2}{(e^{x}-1-ln{(1+x)})^3} overset{??}{=} frac{1}{36} $$

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!

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 $$ lim_{x to 0} frac{(sin x - tanh x)^2}{(e^{x}-1-ln{(1+x)})^3}=lim_{x to 0} frac{bigg[(x-frac16x^3)-(x-frac13x^3)+O(x^5)bigg]^2}{bigg[(x+frac12x^2+frac16x^3)-(x-frac12x^2+frac13x^3)+O(x^4)bigg]^3}=frac{frac{1}{6^2}}{ 1}= frac{1}{36}. $$

Correct answer by xpaul on December 25, 2020

$$L=lim_{x to 0} frac{(sin x - tanh x)^2}{(e^{x}-1-ln{(1+x)})^3} overset{??}{=} frac{1}{36}$$ 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 $$L=lim_{xto 0} frac{x^6/(36)}{x^6}=frac{1}{36}.$$

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