Question

$\lim_{x \to \ 0} \frac{In(1+sinx)}{In 1+sin2x} =$
.................

Answer 1

limita din ln(1+ux)/ux=1
 \lim_{n \to \ 0} \frac{ \frac{ln(1+sinx)}{sinx}*sinx }{sin2x}=[/tex]
lim sin(ux)/ux=1
[tex]= \lim_{n \to \ 0} \frac{sinx}{sin2x} = \frac{0}{0} = \\ \\ [/tex]
$= \lim_{n \to \ 0} \frac{ \frac{sinx}{x} *x}{ \frac{sin2x}{2x}*2x }= \frac{1}{2}$