Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to0}\frac{\ln(1+x)}{\cos x+e^x-1}=0$$
$$A=\lim_{x\to0}\frac{\ln(1+x)}{\cos x+e^x-1}$$ As $x$ approaches $0$, $\ln(1+x)$ approaches $\ln(1+0)=\ln1=0$ and $(\cos x+e^x-1)$ approaches $\cos0+e^0-1=1+1-1=1$ So this is not an indeterminate form, but another with which we can replace $x=0$ right away. $$A=\frac{\ln(1+0)}{\cos0+e^0-1}$$ $$A=\frac{0}{1}$$ $$A=0$$