lim-x-0-ln-sin-2x-ln-sin-4x-

Question Number 91954 by john santu last updated on 04/May/20

\lim_{x \to 0^{+}} \frac{\ln (\sin 2 x)}{\ln (\sin 4 x)}

Commented by jagoll last updated on 04/May/20

{doesn}^{'} t exist

Commented by jagoll last updated on 04/May/20

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Commented by abdomathmax last updated on 04/May/20

h o s p i t a l t h e o r e m

{l i m}_{x \to 0^{+}} \frac{l n (s i n (2 x))}{l n (s i n (4 x))} = {l i m}_{x \to 0^{+}} \frac{\frac{2 c o s (2 x)}{s i n (2 x)}}{\frac{4 c o s (4 x)}{s i n (4 x)}}

= {l i m}_{x \to 0^{+}} \frac{2 c o s (2 x)}{s i n (2 x)} \times \frac{s i n (4 x)}{4 c o s (4 x)}

= {l i m}_{x \to 0^{+}} \frac{1}{2} \frac{c o s (2 x)}{c o s (4 x)} \times \frac{s i n (4 x)}{s i n (2 x)}

= \frac{1}{2} \times 1 \times 2 = 1

Commented by jagoll last updated on 04/May/20

yes ��