I'm trying to solve the following limit:
$$\lim_{x\rightarrow0}\frac{\log(1+\sin{x})-\log(1+x)}{x-\tan{x}}$$
It is pretty straightforward by substituting those expressions by their Taylors polynomial:
Let $y:=\sin{x}$, then that limit is:
$$\lim_{x\rightarrow0}\frac{\log(1+\sin{x})-\log(1+x)}{x-\tan{x}}=\lim_{x\rightarrow0}\frac{\log(1+y)-\log(1+x)}{x-\tan{x}}=\lim_{x\rightarrow0}\frac{[y+o(y)]-[x+o(x)]}{x-[x+\frac{-x^3}{3}+o(x^3)]}=\lim_{x\rightarrow0}\frac{[[x-\frac{x^3}{6}+o(x^3)]+o(x)]-[x+o(x)]}{x-[x+\frac{-x^3}{3}+o(x^3)]}=\lim_{x\rightarrow0}\frac{\frac{-x^3}{6}+o(x)+o(x^3)}{\frac{-x^3}{3}+o(x^3)}$$
Now, that seems to be $\frac{1}{2}$, and it is indeed, but I think I'm doing something wrong with how I manipulate Landau's notation, because I don't know how to simplify that to get the $\frac{1}{2}$. I'm having this same problem in several other limits, I'm not sure whether I'm using it properly or not, any help on this subject would be greatly appreciated.
Here is another simpler method of doing it using L'hopitals rule thrice.
enjoy!
Your result is correct but your method isn't perfect since you should write the taylor series at the order $3$. Let's show how to proceed for the numerator $$\log(1+\sin x)-\log(1+x)=\log\left(1+x-\frac{x^3}6+o(x^3)\right)-x+\frac{x^2}2-\frac{x^3}3+o(x^3)\\=\left(x-\frac{x^3}6\right)-\frac{\left(x-\frac{x^3}6\right)^2}{2}+\frac{\left(x-\frac{x^3}6\right)^3}{3}-x+\frac{x^2}2-\frac{x^3}3+o(x^3)\\=\left(x-\frac{x^3}6\right)-\frac{x^2}{2}+\frac{x^3}{3}-x+\frac{x^2}2-\frac{x^3}3+o(x^3)=-\frac{x^3}6+o(x^3)$$ and your work for the denominator is correct.