Question-201680

Question Number 201680 by cortano12 last updated on 10/Dec/23

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Answered by Calculusboy last updated on 11/Dec/23

S o l u t i o n : s u b s t i t u t e d i t e c t l y, w e g e t \frac{0}{0} (i n d e t e r m i n a n t)

l e t p = 2 s i n x - s i m 2 x \frac{d p}{d x} = 2 c o s x - 2 c o s 2 x

l e t q = s i n x - x c o s x \frac{d q}{d x} = c o s x - (c o s x - x s i n x)

l i \underset{x \to 0}{m} \frac{p}{q} = l i \underset{x \to 0}{m} \frac{2 c o s x - 2 c o s (2 x)}{c o s x - c o s x + x s i n x} = l i \underset{x \to 0}{m} \frac{2 c o s x - 2 c o s (2 x)}{x s i n x}

\frac{l i \underset{x \to 0}{m} (2 c o s x - 2 c o s (2 x))}{l i \underset{x \to 0}{m} (x s i n x)} = \frac{0}{0}

l e t u = 2 c o s x - 2 c o s (2 x) \frac{d u}{d x} = - 2 s i n x + 4 s i n (2 x)

l e t k = x s i n x \frac{d k}{d x} = s i n x + x c o s x

l i \underset{x \to 0}{m} \frac{u}{k} = l i \underset{x \to 0}{m} \frac{4 s i n (2 x) - 2 s i n x}{s i n x + x c o s x} = \frac{0}{0}

l e t j = 4 s i n (2 x) - 2 s i n x \frac{d j}{d x} = 8 c o s (2 x) - 2 c o s x

l e t g = s i n x + x c o s x \frac{d g}{d x} = c o s x + (c o s x - x s i n c) = 2 c o s x - x s i n x

l i \underset{x \to 0}{m} \frac{j}{g} = l i \underset{x \to 0}{m} \frac{8 c o s (2 x) - 2 c o s x}{2 c o s x - x s i n x} = \frac{8 - 2}{2 - 0} = \frac{6}{2} = 3

∴ l i \underset{x \to 0}{m} \frac{2 s i n x - s i n (2 x)}{s i n x - x c o s x} = 3