Question

limit x tends to 0 (ln sin3x / ln sinx) equals to

Answer 1

this is ur required result

Answer 2

Answer:

Step-by-step explanation:

The given equation is:

$\lim_{x\rightarrow 0}\frac{ln(sin3x)}{ln(sinx)}$

Applying the L'hospital rule, we get

=$\lim_{x\rightarrow 0}\frac{\frac{1}{sinx}{\times}cos3x{\times}3}{\frac{1}{sinx}{\times}cosx{\times}1}$

=$\lim_{x\rightarrow 0}\frac{3cos3xsinx}{cosxsin3x}$

Again Applying the L'hospital rule, we get

=$\lim_{x\rightarrow 0}\frac{3(cos3xcosx-sinxsin3x)}{3cosxcos3x-sinxsin3x}$

=$\frac{3(1-0)}{3(1)}=1$