Question

Compute the limit 2Be%5E%7Bx%7D%20-1%7D" id="TexFormula1" title="\lim_{n \to \ 0} \frac{log(1+x)}{cos(x)+e^{x} -1}" alt="\lim_{n \to \ 0} \frac{log(1+x)}{cos(x)+e^{x} -1}" align="absmiddle" class="latex-formula">

Answer 1

I assume you mean

$\displaystyle \lim_{x\to0} \frac{\log(1+x)}{\cos(x) + e^x - 1}$

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

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