Compute $\lim_{x \rightarrow 0} \frac{1-\arctan (\sin(x)+1)}{e^{x}-1}$

You can use the L'Hôpital's rule. taking the derivative of the numerator and the denominator we get: 
\[\lim_{x \rightarrow 0} \frac{1-\arctan (\sin x+1)}{e^{x}-1}=\lim_{x \rightarrow 0} \frac{-\frac{\cos x}{1+(1+\sin x)^2}}{e^{x}}= \frac{-\frac{\cos 0}{1+(1+\sin 0)^2}}{e^{0}}=-\frac{1}{2}. \]