<s>[INST] help determining $a,b$ such that limit will exist and calculating it So i have a limit: 
$$\lim_{x\rightarrow 0} \frac{e^{ax}+b\ln(2-x)-1}{x-\sin{x}}$$
And i need to determine values for $a,b$ so that the limit will exist, and calculate the limit then.
There is also a $hint$, that i need to use Taylor series for this, so it might be helpful, but i don't know if i should look for the whole function or just a part of fraction.
Any help would really be appreciated.
Thank you in advance.
 [/INST] It is sure that Taylor series help a lot. Let us try
$$e^{ax}=1+a x+\frac{a^2 x^2}{2}+\frac{a^3 x^3}{6}+O\left(x^4\right)$$ $$\log(x-2)=\log (2)-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{24}+O\left(x^4\right)$$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ Replacing $$\frac{e^{ax}+b\ln(2-x)-1}{x-\sin{x}}=\frac{b \log (2)+x \left(a-\frac{b}{2}\right)+x^2
   \left(\frac{a^2}{2}-\frac{b}{8}\right)+x^3
   \left(\frac{a^3}{6}-\frac{b}{24}\right)+O\left(x^4\right)}{\frac{x^3}{6}+O\left(x^4\right)}$$ In order the limit exists, the constant term and the coefficients of the first and second powers of $x$ in numerator must be $0$. Then $???$
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