doubts asking website

Is there any website where we can ask doubts for JEE and they answer soon for free or for money less than 100 rs??

2 Answers

1
a1b2c3 shivam jain ·

evaluate limit x tends to '0' (sin(pi(cos square x)/x square

1
johncenaiit ·

@ Athenes .... i know many websites... but target iit is best (for jee)

@a1b2c3,
limx0x2sin(πcos2x)=limx0x2sin(ππcos2x)=limx0x2sin(πsin2x)×πsin2xπsin2x

Ans : π

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