Question

Need a hint on how to solve lim(x->0)(ln(1+3x)/sinx).
(Without l'hopitals.)

Answer 1

Initial thoughts are substituting t=arcsinx, t->0 when x->0, x=sint,
I get stuck at ln(1+3sint)/t though; could you rewrite this somehow to ln(x+1)/x?

Answer 2

\[\lim_{x\to0} \frac{\ln{(1+3x)}}{\sin{x}}=\lim_{x\to0} \frac{\ln(1+3x)}{x}\cdot \frac{x}{\sin{x}}=\lim_{x\to0} \frac{\ln(1+3x)}{x}\cdot \lim_{x\to0}\frac{x}{\sin{x}}\]
Substituting u=3x on the first one should make the answer clear.