Answer
$$
\lim _{x \rightarrow 0} \frac{\ln (1+x)}{\cos x+e^{x}-1}=0
$$
Work Step by Step
$$
\lim _{x \rightarrow 0} \frac{\ln (1+x)}{\cos x+e^{x}-1}
$$
The required limit is, in fact, easy to find because the function is continuous at $0$ and the denominator is nonzero there:
$$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\ln (1+x)}{\cos x+e^{x}-1} &=\frac{\ln 1}{1+1-1} \\
&=\frac{0}{1}\\
&= 0
\end{aligned}
$$
