I assume you mean
[tex]\displaystyle \lim_{x\to0} \frac{\log(1+x)}{\cos(x) + e^x - 1}[/tex]
since the limand is free of n. As x goes to 0, the numerator converges to log(1 + 0) = 0, while the denominator converges to cos(0) + e⁰ - 1 = 1, so the overall limit is
[tex]\displaystyle \lim_{x\to0} \frac{\log(1+x)}{\cos(x) + e^x - 1} = \frac01 = \boxed{0}[/tex]
