limits again

Evaluate the limit.

lim(x→0) √(1-cos2x)/sinx

5 Answers

21
tapanmast Vora ·

is it 0???

1
maydayhay ·

how?

1
? ·

guys !! .limit does not exist !

106
Asish Mahapatra ·

yup unique u r on spot.
(1-cos2x)/sinx
=√2sinx
for x→0-, it is square root of negative no. which does not exist.
hence LHL does not exist
hence limit does not exist.

21
tapanmast Vora ·

OHHHH
THAT WAS A BLUNDER!!!!!1

YEAH!!!

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