$$\lim \limits_{x \to 0} \frac{\sin(x)-x}{\log(1+x)-1-2x+e^x}$$
I've tried with equivalents of $\sin(x)=x; \log(1+x)=x$ and l'Hopital, can someone give me hints?
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try $$ sinx=x- \frac{x^3}{3!}\e^x=1+x+\frac{x^2}{2!}+..\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-...$$ – Khosrotash Apr 10 '15 at 15:21
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Use l'Hopital several times.
$$\begin{align}\lim_{x\to 0}\frac{\sin x -x}{\ln (x+1)-1-2x+e^x}&=\lim_{x\to 0}\frac{\cos x-1}{\frac{1}{x+1}-2+e^x}\\&=\lim_{x\to 0}\frac{-\sin x}{-\frac{1}{(x+1)^2}+e^x}\\&=\lim_{x\to 0}\frac{-\cos x}{\frac{2}{(x+1)^3}+e^x}\\&=-\frac 13.\end{align}$$
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