210k views
25 votes
Compute the limit
\lim_(n \to \ 0) (log(1+x))/(cos(x)+e^(x) -1)

1 Answer

2 votes

I assume you mean


\displaystyle \lim_(x\to0) (\log(1+x))/(\cos(x) + e^x - 1)

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


\displaystyle \lim_(x\to0) (\log(1+x))/(\cos(x) + e^x - 1) = \frac01 = \boxed{0}

User Ugurerkan
by
4.9k points