Find: \(\lim\limits_{x\to 0}\) x log sin2 x

1.	Find: \(\lim\limits_{x\to 0}\) x log sin2 x
Answer» \(\lim\limits_{x\to 0}\) x log sin² x (0 x \(\infty\) type) = \(\lim\limits_{x\to 0}\) \(\frac{log sin^2x}{1/x}\) (\(\infty\)/\(\infty\) type) = \(\lim\limits_{x\to 0}\) \(\cfrac{\frac{2sinxcosx}{sin^2x}}{-\frac1{x^2}}\) (By D.L.H. Rule) = \(\lim\limits_{x\to 0}\) \(\frac{-2x^2}{sin^2x}.sin x.cos x\) = -2(\(\lim\limits_{x\to 0}\) \(\frac{x}{sin x}\))² . \(\lim\limits_{x\to 0}\) sin x cos x = -2 .1². 0 . 1 (\(\because\lim\limits_{x\to 0}\frac{x}{sin x}=1\)) = 0 \(\therefore\) \(\lim\limits_{x\to 0}\) x log sin²x = 0

Answer»

\(\lim\limits_{x\to 0}\) x log sin² x (0 x \(\infty\) type)

= \(\lim\limits_{x\to 0}\) \(\frac{log sin^2x}{1/x}\) (\(\infty\)/\(\infty\) type)

= \(\lim\limits_{x\to 0}\) \(\cfrac{\frac{2sinxcosx}{sin^2x}}{-\frac1{x^2}}\) (By D.L.H. Rule)

= \(\lim\limits_{x\to 0}\) \(\frac{-2x^2}{sin^2x}.sin x.cos x\)

= -2(\(\lim\limits_{x\to 0}\) \(\frac{x}{sin x}\))² . \(\lim\limits_{x\to 0}\) sin x cos x

= -2 .1². 0 . 1 (\(\because\lim\limits_{x\to 0}\frac{x}{sin x}=1\))

= 0

\(\therefore\) \(\lim\limits_{x\to 0}\) x log sin²x = 0

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