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Question Number 60386 by Kunal12588 last updated on 20/May/19

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

w h y i t c a n n o t b e s o l v e d t h i s w a y

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

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

= π \times \underset{x \to 0}{l i m} \frac{{c o s}^{2} x}{x^{2}}

b u t i t i s n o t e q u a l t o π

Commented by mr W last updated on 20/May/19

\underset{x \to 0}{l i m} \frac{s i n (π {c o s}^{2} x)}{π {c o s}^{2} x} = \frac{\sin π}{π \times 1} = \frac{0}{π} = 0 \neq π

\lim_{x \to 0} \frac{π {c o s}^{2} x}{x^{2}} = \frac{π}{0} \to \infty

Commented by mr W last updated on 20/May/19

o r

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

= \underset{x \to 0}{l i m} \frac{\cos (π {c o s}^{2} x) (- π 2 \cos x \sin x)}{2 x}

= \underset{x \to 0}{l i m} \frac{(- 1) (- π) \sin x}{x}

= π \underset{x \to 0}{l i m} \frac{\sin x}{x}

= π

Commented by prakash jain last updated on 20/May/19

Limit of product of two function = product of limits this is only true if both limit exist and finite.

Commented by Mr X pcx last updated on 20/May/19

a n o t h e r w a y w e h s v e c o s x \sim 1 - \frac{x^{2}}{2} \Rightarrow

{c o s}^{2} x \sim 1 - x^{2} + \frac{x^{4}}{4} \sim 1 - x^{2} + o (x^{4}) \Rightarrow

π {c o s}^{2} x \sim π - π x^{2} \Rightarrow \frac{s i n (π {c o s}^{2} x)}{x^{2}}

\sim \frac{s i n (π x^{2})}{x^{2}} \sim \frac{π x^{2}}{x^{2}} (= π) \Rightarrow

{l i m}_{x - 0} \frac{s i n (π {c o s}^{2} x)}{\overset{2}{x}} = π .

Answered by mr W last updated on 20/May/19

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

= \underset{x \to 0}{l i m} \frac{s i n (π - π \sin^{2} x)}{x^{2}}

= \underset{x \to 0}{l i m} \frac{s i n (π \sin^{2} x)}{x^{2}}

= \underset{x \to 0}{l i m} {\frac{s i n (π \sin^{2} x)}{π \sin^{2} x} \times π \times {(\frac{\sin x}{x})}^{2}}

= 1 \times π \times 1^{2}

= π

Commented by Prithwish sen last updated on 20/May/19

Sir I think it will be

\lim_{x \to 0} \frac{\sin (π - π \cos^{2} x)}{x^{2}} (2^{nd} line)

please check .

Commented by Kunal12588 last updated on 20/May/19

y e s s i r i k n o w t h e a n s w e r t h i s w a y b u t

w h y i t c a n n o t b e s o l v e d t h e w a y i h a v e w r i t t e n

Commented by mr W last updated on 20/May/19

\cos^{2} x = 1 - \sin^{2} x

\Rightarrow π \cos^{2} x = π - π \sin^{2} x

\Rightarrow \sin (π \cos^{2} x) = \sin (π - π \sin^{2} x)

Commented by Prithwish sen last updated on 21/May/19

Yes sir you are right . Please forgive .

Commented by mr W last updated on 22/May/19

n o p r o b l e m s i r!

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