Is there any website where we can ask doubts for JEE and they answer soon for free or for money less than 100 rs??
-
UP 0 DOWN 0 0 2
2 Answers
1evaluate limit x tends to '0' (sin(pi(cos square x)/x square
1@ Athenes .... i know many websites... but target iit is best (for jee)
@a1b2c3,
\lim_{x\rightarrow 0}\frac{sin(\pi cos^{2}x)}{x^{2}}=\lim_{x\rightarrow 0}\frac{sin(\pi -\pi cos^{2}x)}{x^{2}}=\lim_{x\rightarrow 0}\frac{sin(\pi sin^{2}x)}{x^{2}}\times \frac{\pi sin^{2}x}{\pi sin^{2}x}
Ans : \pi
